## How to solve exponential equations

Examples of exponential equations.

$$ 2^{\red x} = 4 \\ 8^{\red{2x}} = 16 \\ \\ 16^{\red { x+1}} = 256 \\ \left( \frac{1}{2} \right)^{\red { x+1}} = 512 $$

As you might've noticed, an exponential equation is just a special type of equation. It's an equation that has exponents that are $$ \red{ variables}$$.

## Steps to Solve

There are different kinds of exponential equations. We will focus on exponential equations that have a single term on both sides. These equations can be classified into 2 types.

$$ 4^x = 4^9 $$.

$$ 4^3 = 2^x $$.

$$ \left( \frac{1}{4} \right)^x = 32 $$ (Part II below)

## Part I. Solving Exponential Equations with Same Base

Solve: $$ 4^{x+1} = 4^9 $$

Ignore the bases, and simply set the exponents equal to each other

$$ x + 1 = 9 $$

Solve for the variable

$$ x = 9 - 1 \\ x = \fbox { 8 } $$

$$ 4^{x+1} = 4^9 \\ 4^{\red{8}+1} = 4^9 $$ $$ 4^{\red{9} } = 4^9 $$

## Exponential Equation Solver

Enter any exponential equation into the algebra solver below :

## II. Solving Exponential Equations with un -like bases

What do they look like.

$$ \red 4^3 = \red 2^x $$ $$ \red 9^x = \red { 81 } $$ $$ \left( \red{\frac{1}{2}} \right)^{ x+1} = \red 4^3 $$ $$ \red 4^{2x} +1 = \red { 65 } $$

In each of these equations, the base is different. Our goal will be to rewrite both sides of the equation so that the base is the same.

Solve: $$ 4^{3} = 2^x $$

Answer : They are both powers of 2

Rewrite equation so that both exponential expressions use the same base

$$ \red 4^{3} = 2^x \\ (\red {2^2})^{3} = 2^x $$

Use exponents laws to simplify

$$ (\red {2^2})^{3} = 2^x \\ (2^\red {2 \cdot 3 }) = 2^x \\ (2^\red 6 ) = 2^x $$

Solve like an exponential equation of like bases

$$ (2^\red 6 ) = 2^x \\ x = \fbox{6} $$

Substitute $$\red 6 $$ into the original equation to verify our work.

$$ 4^{3} = 2^{\red 6} $$ $$ 64 = 64 $$

## Example with Negative Exponent

Unlike bases often involve negative or fractional bases like the example below. We are going to treat these problems like any other exponential equation with different bases--by converting the bases to be the same.

## Practice Problems ( un -like bases)

Solve the following exponential Equation: $$9^x = 81$$

You can use either 3 or 9 . I will use 9.

$ \\ 81 = \red 9 ^{\blue 2} \\ 9 = \red 9 ^{\blue 1} \\ $

Substitute the rewritten bases into original equation

$$ (\red 9^{\blue 1})^x = \red 9^{\blue 2} $$ 2 ) x =3 4 -->

$$ (\red 9^{\blue 1})^x = \red 9^{\blue 2} \\ 9^{1 \cdot x } = 9 ^{2} \\ 9^{x } = 9 ^{2} $$ 2 x =3 4 -->

$$ x = 2 $$ -->

Solve the equation : $$ 4^{2x} +1 = 65 $$

Rewrite this equation so that it looks like the other ones we solved. Isolate the exponential expression as follows:

$$ 4^{2x} +1 \red{-1} = 65\red{-1} \\ 4^{2x} = 64 $$

They are both powers of 2 and of 4 . You could use either base to solve this. I will use base 4

$ \\ 64 = \red 4 ^{\blue 3} \\ 4 = \red 4 ^{\blue 1} \\ $

$$ 4^{2x} = 64 \\ \red 4^{\blue{ 2x }} = \red 4^{\blue 3 } $$

$$ \red 4^{\blue{ 2x }} = \red 4^{\blue 3 } $$ 2x = 4 3 -->

Not much to do this time :)

$$ 2x = 3 \\ x = \frac{3}{2} $$

Solve the exponential Equation : $$ \left( \frac{1}{4} \right)^x = 32 $$

Since these equations have different bases, follow the steps for unlike bases

They are both powers of 2

$ \\ 32 = \red 2 ^{\blue 5} \\ \frac 1 4 = \red 2 ^{\blue {-2}} \\ $

Rewrite as a negative exponent and substitute the rewritten bases into original equation

$$ \left( \frac{1}{4} \right)^x = 32 \\ \left( \frac{1}{2^2} \right)^x = 32 \\ \left(\red 2 ^{\blue{-2}} \right)^x = \red 2^{\blue 5} $$

$$ \left(\red 2 ^{\blue{-2}} \right)^x = \red 2^{\blue 5} \\ 2 ^{-2 \cdot x} = 2^5 \\ 2 ^{-2x} = 2^5 $$

$$ 2 ^{-2x} = 2^5 \\ -2x = 5 \\ \frac{-2x}{-2} = \frac{5}{-2} \\ x = -\frac{5}{2} $$

Solve this exponential equation: $$ \left( \frac{1}{9} \right)^x-3 = 24 $$

$$ \left( \frac{1}{9} \right)^x -3 \red{+3} =24\red{+3} \\ \left( \frac{1}{9} \right)^x=27 $$

They are both powers of 3 .

$ \\ \frac 1 9 = \red 3 ^{\blue {-2}} \\ 27 = \red 3 ^{\blue 3} \\ $

Rewrite as a negative exponent and substitute into original equation

$$ \left( \frac{1}{9} \right)^x=27 \\ \left( \red{3^{-2}}\right)^x=\red{3^3 } $$

$$ 3^\red{{-2 \cdot x}} = 3^3 \\ 3^\red{{-2x}} = 3^3 $$

$$ -2x = 3 \\ x = \frac{3}{-2} \\ x = -\frac{3}{2} $$

Solve this exponential equation: $$ \left( \frac{1}{25} \right)^{(3x -4)} -1 = 124 $$

Rewrite this equation so that it looks like the other ones we solved--In other words, isolate the exponential expression as follows:

$$ \left( \frac{1}{25} \right)^{(3x -4)} -1 \red{+1} = 124 \red{+1} \\ \left( \frac{1}{25} \right)^{(3x -4)} = 125 $$

They are both powers of 5 .

$ \\ \frac { 1 } { 25 } = \red 5 ^{\blue {-2}} \\ 125 = \red 5 ^{\blue 3} \\ $

$$ \left( \frac{1}{25} \right)^{(3x -4)} = 125 \\ \left( \red{5^{-2}} \right)^{(3x -4)} = \red{5^3} $$

$$ 5^\red{{-2 \cdot (3x -4)}} = 5^3 \\ 5^\red{{(-6x + 8)}} = 5^3 $$

$$ -6x + 8 =3 \\ -6x = -5 \\ x = \frac{-5}{-6} \\ x = \frac{5}{6} $$

- Laws of Exponents
- Rules of Exponents

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## How to Solve Exponential Equations

Last Updated: November 25, 2022 Fact Checked

This article was co-authored by David Jia . David Jia is an Academic Tutor and the Founder of LA Math Tutoring, a private tutoring company based in Los Angeles, California. With over 10 years of teaching experience, David works with students of all ages and grades in various subjects, as well as college admissions counseling and test preparation for the SAT, ACT, ISEE, and more. After attaining a perfect 800 math score and a 690 English score on the SAT, David was awarded the Dickinson Scholarship from the University of Miami, where he graduated with a Bachelor’s degree in Business Administration. Additionally, David has worked as an instructor for online videos for textbook companies such as Larson Texts, Big Ideas Learning, and Big Ideas Math. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 139,035 times.

Exponential equations may look intimidating, but solving them requires only basic algebra skills. Equations with exponents that have the same base can be solved quickly. In other instances, it is necessary to use logs to solve. Even this method, however, is simple with the aid of a scientific calculator.

## Equating Two Exponents with the Same Base

## Equating an Exponent and a Whole Number

## Using Logs for Terms without the Same Base

## Expert Q&A

## You Might Also Like

- ↑ https://www.mathsisfun.com/definitions/exponent.html
- ↑ http://www.mathwarehouse.com/algebra/exponents/solve-exponential-equations-how-to.php
- ↑ https://courses.lumenlearning.com/intermediatealgebra/chapter/use-compound-interest-formulas/
- ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut45_expeq.htm
- ↑ http://www.purplemath.com/modules/solvexpo.htm
- ↑ http://www.mathsisfun.com/algebra/exponents-logarithms.html
- ↑ https://tutorial.math.lamar.edu/classes/alg/solveexpeqns.aspx
- ↑ https://www.khanacademy.org/math/algebra2/x2ec2f6f830c9fb89:logs/x2ec2f6f830c9fb89:exp-eq-log/a/solving-exponential-equations-with-logarithms

## About This Article

To solve exponential equations with the same base, which is the big number in an exponential expression, start by rewriting the equation without the bases so you're left with just the exponents. Then, solve the new equation by isolating the variable on one side. To check your work, plug your answer into the original equation, and solve the equation to see if the two sides are equal. If they are, your answer is correct. To learn how to solve exponential equations with different bases, scroll down! Did this summary help you? Yes No

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## How to Solve Exponential Equations?

An exponential equation is an equation with exponents in which exponents (or) is part of the exponents is variable. Here, you learn more about solving exponential equations problems.

When the power is variable and if it is part of an equation, it is called an exponential equation . It may be necessary to use the relationship between power and logarithm to solve the exponential equations.

## Related Topics

- How to Solve Rational Exponents
- How to Solve Rational Exponents and Radicals

## Step-by-step guide to exponential equations

There are three types of exponential equations. They are as follows:

- Equations with the same bases on both sides
- Equations with different bases can be made the same
- Equations with different bases that cannot be made the same

## Exponential equations formulas

When solving an exponential equation, the bases of both sides may be the same or may not be the same. Here are the formulas used in each of these cases.

## Solving exponential equations with the same bases

Sometimes, even though the exponents of both sides are not the same, they can be made the same. To solve exponential equations in each of these cases, we use only the property of equality of exponential equations, by which we equalize the exponentials and solve for the variable.

## Solving exponential equations with different bases

Sometimes the bases on either side of an exponential equation may not be the same (or) cannot be made the same. We solve exponential equations using logarithms when the bases on both sides of the equation are not the same. In such cases, we can do one of the following:

- Convert the exponential equation into the logarithmic form using the formula \(b^x=a⇔log _b\left(a\right)=x\)
- Apply \(log\) on both sides of the equation and solve for the variable. In this case, we have to use the logarithm property.

## Exponential Equations – Example 1:

solve the equation \(7^x=3\).

The bases on both sides of the exponential equation are not the same, so must apply \(log\) on both sides of the exponential equation:

\(log 7^x=log 3\)

Then, use the property of \(log\): \(log a^m=m \:log a\)

\(x log 7=log 3\)

Now, dividing both sides by \(log 7\):

\(x=\frac{log 3}{log 7}\)

## Exponential Equations – Example 2:

Solve the equation \(4^{2x-1}=64\).

The bases are not the same, but we can rewrite \(64\) as a base of \(4\) → \(4^3=64\)

Then, rewrite the equation as:

\(4^{2x-1}=4^3\)

With the property of exponential functions, if the bases are the same, the exponents must be equal:

\(2x-1=3 → 2x=3+1 → 2x=4\)

Now, divide each side by \(2\):

## Exercises for Exponential Equations

Solve exponential equations..

- \(\color{blue}{\frac{81}{3^{-x}}=3^6}\)
- \(\color{blue}{5^{3x-2}=125^{2x}}\)
- \(\color{blue}{5^{2x}=21}\)
- \(\color{blue}{4^{x-2}=0.125}\)
- \(\color{blue}{3^{^{2x+1}}=15}\)
- \(\color{blue}{x=2}\)
- \(\color{blue}{x=-\frac{2}{3}}\)
- \(\color{blue}{x=\frac{log\:21}{2\:log\:5}}\)
- \(\color{blue}{x=\frac{1}{2}}\)
- \(\color{blue}{x=\frac{log\:5}{2\:log\:3}}\)

by: Effortless Math Team about 2 years ago (category: Articles )

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## How to Solve Exponents: A Step-by-Step Guide with Examples

Exponents, also known as powers or indices, are a fundamental concept in mathematics. It plays a crucial role in various fields, from basic arithmetic to advanced calculus. An exponent represents how many times a number, called the base, is multiplied by itself. To solve exponents involves understanding the properties of exponents, simplifying expressions, and performing operations with exponential terms. In this guide, we will walk you through the process of solving exponent problems using maths.ai as a teaching tool.

## Understanding Exponents

Before we delve into solving exponents problems, let’s review the basics of exponents.

An exponent is usually written as a superscript number next to the base. For example, in the expression 2^3, 2 is the base, and 3 is the exponent. This means that 2 is multiplied by itself three times: 2 * 2 * 2 = 8.

## Example 1: Simple Exponentiation

Let’s solve the problem: 4^2.

Step 1: Identify the base and the exponent.

Exponent: 2

Step 2: Apply the exponentiation rule.

4^2 = 4 * 4 = 16

## Exponent Properties

Understanding exponent properties is essential for solving more complex exponent problems.

## Property 1: Product of Powers

a^m * a^n = a^(m + n)

This property states that when you multiply two powers with the same base, you can add their exponents.

## Example 2: Product of Powers

Solve the problem: 3^4 * 3^2.

Step 1: Identify the base and the exponents.

Exponent 1: 4

Exponent 2: 2

Step 2: Apply the product of powers property.

3^4 * 3^2 = 3^(4 + 2) = 3^6 = 729

## Property 2: Quotient of Powers

a^m / a^n = a^(m – n)

This property states that when you divide two powers with the same base, you can subtract their exponents.

## Example 3: Quotient of Powers

Solve the problem: 5^7 / 5^4.

Exponent 1: 7

Exponent 2: 4

Step 2: Apply the quotient of powers property.

5^7 / 5^4 = 5^(7 – 4) = 5^3 = 125

## Property 3: Power of a Power

(a^m)^n = a^(m * n)

This property states that when you raise a power to another power, you can multiply the exponents.

## Example 4: Power of a Power

Solve the problem: (2^3)^4.

Exponent: 4

Step 2: Apply the power of a power property.

(2^3)^4 = 2^(3 * 4) = 2^12 = 4096

## Simplifying Exponential Expressions

Simplifying exponential expressions involves applying the properties of exponents to combine or rewrite terms for easier computation.

## Example 5: Simplifying Exponential Expressions

Simplify the expression: 2^5 * 2^8 / 2^3.

Exponent 1: 5, Exponent 2: 8, Exponent 3: 3

Step 2: Apply the properties of exponents.

2^5 * 2^8 / 2^3 = 2^(5 + 8 – 3) = 2^10 = 1024

## Operations with Exponential Terms

Performing operations with exponential terms involves applying the properties of exponents while considering addition, subtraction, multiplication, and division.

## Example 6: Operations with Exponential Terms

Solve the expression: (4^3 * 4^2) / 4^4 + 4^1.

Step 1: Identify the base and the exponents for each term.

Term 1: 4^3 * 4^2,Term 2: 4^4, Term 3: 4^1

Step 2: Simplify each term using the properties of exponents.

Term 1: 4^3 * 4^2 = 4^(3 + 2) = 4^5, Term 2: 4^4, Term 3: 4^1

Step 3: Combine the terms using addition and subtraction.

(4^5 * 4^4) / 4^4 + 4^1 = 4^5 + 4^1

Step 4: Calculate the final result.

4^5 + 4^1 = 1024 + 4 = 1028

Exponents are a fundamental concept in mathematics that allows us to represent repeated multiplication in a concise form. By understanding the properties of exponents, simplifying expressions, and performing operations with exponential terms, you can confidently tackle various exponent problems. Remember to apply the rules systematically, and when in doubt, use tools maths.ai to assist you in your journey towards mastering exponents. With practice and a clear grasp of exponent principles, you’ll be well-equipped to solve more complex maths problems that involve exponents.

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## Exponential and logarithmic function

5.1 EXPONENTIAL FUNCTIONS

Recall from Chapter 1 the deﬁnition of a^r , where r is a rational number: if r = m/n , then for appropriate values of m and n ,

a^(m/n)=(root(n,a))^m

For example

16^(3/4)=(root(4,16))^3=2^3=8 ,

27^-(1/3)=1/(27^(1/3))=1/(root(3,27))=1/3 ,

and 64^-(1/2)=1/(64^(1/2))=1/(root(64))=1/8 .

In this section the deﬁnition of a^r is extended to include all real (not just rational) values of the exponent r . For example, the new symbol 2^(root(3) might be evaluated by approximating the exponent root(3) by the numbers 1.7,1.73,1.732 . and so on. Since these decimals approach the value of root(3) more and more closely, it seems reasonable that 2^(root(3) should be approximated more and more closely by the numbers to 2^(1.7),2^(1.73),2^(1.732) , and so on. (Recall, for example. that 2^(1.7)=2^(17/10)=root(10,2^17) ) In fact. this is exactly how 2^(root(3) is deﬁned (in a more advanced course).

With this interpretation of real exponents, all rules and theorems for exponents are valid for real-number exponents as well as rational ones. In addition to the rules for exponents presented earlier, several new properties are used in this chapter. For example, if y=2^x , then each real value of x leads to exactly one value of y , and therefore. y=2^x deﬁnes a function. Furthermore,

if 3^x=3^4 , then x=4 ,

and for p>0 ,

if p^2=3^2 , then p=3 .

Also, 4^2<4^3 but (1/2)^2>(1/2)^3 ,

so that when a > 1 , increasing the exponent on a leads to a larger number, but if 0 < a < 1 , increasing the exponent on a leads to a smaller number.

These properties are generalized below. Proofs of the properties are not given here, as they require more advanced mathematics.

ADDITIONAL PROPERTIES OF EXPONENTS

(a) If a > 0 and a!=1 , then a^x is a unique real number for all real numbers x . (b) In a>0 and a!=1 , then a^b=a^c if and only if b=c .

(c) If a > 1 and m < n , then a^m<a^n .

(d) If 0 < a < 1 and m < n , then a^m>a^n .

Properties (a) and (b) require a>0 so that a^x is always deﬁned. For example, (-6)^x is not a real number if x = 1/2 . This means that a^x will always he positive, since a is positive. In part (a), a!=1 because 1^x=1 for every real-number value of x , so that each value of x does not lead to a distinct real number. For Property (b) to hold, a must not equal 1 since, for example. 1^4=1^5 , even though 4!=5 .

EXPONENTIAL EQUATIONS The properties given above are useful in solving equations, as shown by the next examples.

USING A PROPERTY OF EXPONENTS TO SOLVE AN EQUATION

Solve (1/3)^x=81 .

First, write 1/3 as 3^-1 , so that (1/3)^x=3^(-x) . Since 81=3^4 ,

(1/3)^x=81

becomes

By the second property above,

-x=4 , or x=-4 .

The solution set of the given equation is {-4} .

In Section 5.4 we describe a more general method for solving exponential equations where the approach used in Example 1 is not possible. For instance, this method could not be used to solve an equation like 7^x=12 , since it is not easy to express both sides as exponential expressions with the same base.

Solve 81=b^(4/3) .

Being by writing b^(4/3) as (root(3,b))^4 .

81=b^(4/3)

81=(root(3,b))^4

+-3=root(3,b) Take fourth roots on both sides.

+-27=b Cubs both sides.

Check both solutions in the original equation. Since both solutions check, the solution set is {-27,27} .

GRAPHING EXPONENTIAL FUNCTIONS As mentioned above, the expression a^x satisﬁes all the properties of exponents from Chapter 1. We can now deﬁne a function f(x) = a^x whose domain is the set of all real numbers (and not just the rationals).

EXPONENTIAL FUNCTION If a>0 and a!=1 , then

f(x) = a^x

deﬁnes the exponential function with base a .

NOTE If a=1 , the function is the constant function f(x) = 1 , and not an exponential function.

EVALUATING AN EXPONENTIAL EXPRESSION

If f(x)=2^x , find each of the following.

(a) f(-1)

Replace x with -1 .

f(-1)=2^-1=1/2

(b) f(3)=2^3=8

(c) f(5/2)=2^(5/2)=(2^5)^(1/2)=32^(1/2)=root(32)=4root(2)

Figure 5.1 shows the graph of f(x)=2^x . The graph was found by obtaining a number of ordered pairs belonging to the function and then drawing a smooth curve through them. As we choose smaller and smaller negative values of x , the y y -values get closer and closer to 0 , as shown in the table below.

FIGURE 5.1

Because 2^x is always positive, the values of y will never become 0 . The line y = 0 . which the graph gets closer and closer to, is called a horizontal asymptote . Asymptotes will be discussed in more detail in Chapter 6. By Property (c), as x increases, so does y , making f(x) = 2^x an increasing function. As suggested by the graph in Figure 5.1, the domain of the function is (-∞,∞) , and the range is (0,∞) .

Figure 5.2

In Figure 5.2, the graph of g(x) = (1/2)^x was sketched in a similar way. The domain and range are the same as those of f(x) = 2^x . However, here, as the values of x increase. the values of y decrease, so g(x) = (1/2)^x is a decreasing function. The graph of g(x) = (1/2)^x is the reﬂection of the graph of f(x) = 2^x across the y -axis, because g(x) = f(-x) .

As the graphs suggest, by the horizontal line test, f(x) = 2^x and g(x) = (1/2)^x are one-to-one functions.

The graph of f(x) = 2^x is typical of graphs of f(x) = a^x . where a>1 . For larger values of a . the graphs rise more steeply, but the general shape is similar to the graph in Figure 5.1. Exponential functions with 0 < a < 1 have graphs similar to that of g(x) = (1/2)^x . Based on our work above, the following generalizations can be made about the graphs of exponential functions deﬁned by f(x) = a^x .

GRAPH OF f(x) = a^x

1. The point (0, 1) is on the graph.

2. If a > 1 , f is an increasing function; if 0 < a < 1 , f is a decreasing function.

3. The x -axis is a horizontal asymptote.

4. The domain is (-∞,∞) and the range is (0,∞) .

We can use the composition of functions lo produce more general exponential functions. If h(u)=ka^u , where k is a constant and u = g(x) , and f(x) = h[g(x) ], then

f(x)=h[g(x)]=ka^(g(x))

For example, if a=7 , g(x)=3x-1 , and k=4 , then

f(x)=4*7^(3x-1)

GRAPHING A COMPOSITE EXPONENTIAL FUNCTION

Graph f(x)=2^(-x+2) The graph will have the same shape as the graph of g(x)=2^(-x)=(1/2)^x . Because of the 2 added to -x , the graph will be translated 2 units to the right, compared with the graph of g(x)=2^(-x) . This means that the point (2,1) is on the graph instead of (0,1) . When x=0 , y=2^2=4 , so the point (0, 4) is on the graph. Plotting a few additional points, such as (-1, 8) and (1. 2) , gives the graph in Figure 5.3. The graph of g(x)=2^(-x) is also shown for comparison.

Figure 5.3

IN SIMPLEST TERMS

The exponential growth of deer in Massachusetts can be calculated using the equation t=50,000(1+0.06)^n , where 50,000 is the initial deer population and .06 is the rate of growth This the total population after n years have passed.

Given the initial population and growth rate above, We could predict the total population after 4 years by using n = 4 .

T=50,000(1+0.06)^4

≈ 50,000(1.26)

= 63,000

We can expect a total population of about 63,000 , or an increase of about 13000 deer.

Graph f(x)=-2^x+3

The graph of y=-2^x is a reﬂection across the x -axis of the graph of y=2^x .The 3 indicates that the graph should be translated up 3 units, as compared to the graph of y=-2^x . Find some ordered pairs. Since y=-2^x would have y -intercept -1 , this function has y -intercept 2 , which is up 3 units from the y-intercept of y=-2^x . Some other ordered pairs are (1,1),(2,-1) , and (3, -5) . For negative values of x , the graph approaches the line y = 3 as a horizontal asymptote. The graph is shown in Figure 5.4.

Figure 2.4

Graph f(x)=2^(-x^2) .

Write f(x)=2^(-x^2) as f(x)=1/(2^x^2) to ﬁnd ordered pairs that belong to the function. Some ordered pairs are shown in the chm below.

As the chant suggests, 0<y<=1 for all values of x . Plotting these points and drawing a smooth curve through them gives the graph in Figure 5.5. This graph is symmetric with respect to the y -axis and has the x -axis as a horizontal asymptote.

Figure 5.5

The important normal curve in probability theory has a graph very similar to me one in Figure 5.5.

COMPOUND INTEREST The formula for compound interest (interest paid on both principal and interest) is an important application of exponential functions. You may recall the formula for simple interest, {Iota}=Prt , where P is the amount left at interest, r is the rate of interest expressed as a decimal, and t is time in years that the principal earns interest. Suppose t=1 year. Then at the end of the year the amount has grown to

P+Pr=P(1+r) ,

the original principal plus the interest. If this amount is left at the same interest rate for another year, the total amount becomes

[P(1+r)]+[P(1+r)]r

= [P(1+r)](1+r)

= P(1+r)^2

After the third year, this will grow to

[P(1+r)^2]+[P(1+r)^2]r

= [P(1+r)^2](1+r)

= P(1+r)^3 .

Continuing in this way produces the following formula for compound interest.

COMPOUND INTEREST

If P dollars is deposited in an account paying an annual rate of interest r compounded (paid) m times per year, then after t years the account will contain A dollars, where

A=P(1+r/m)^(tm) .

For example, let $ 1000 be deposited in an account paying 8 % per year compounded quarterly, or four times per year. After 10 years the account will contain

P(1+r/m)^(tm)

= 1000(1+(0.08)/4)^(10(4))

= 1000(1+0.02)^(40)

= 1000(1.02)^40

dollars. The number (1.02)^40 can be found by using a calculator with a y^x key. To ﬁve decimal places, (1.02)^40=2.20804 . The amount on deposit after 10 years is

1000(1.02)^40=1000(2.20804)=2208.04 ,

or $ 2208.04

In the formula for compound interest. A is sometimes called the future value and P the present value.

FINDING PRESENT VALUE

An accountant wants to buy a new computer in three years that will cost $ 20,000 .

(a) How much should be deposited now, at 6% interest compounded annually, to give the required $ 20,000 in three years?

Since the money deposited should amount to $ 20,000 in three years, $ 20,000 is the future value of the money. To ﬁnd the present value P of $ 20,000 (the amount to deposit now), use the compound interest formula with A = 20,000 , r=0.06 , m=1 ,and t=3 .

A=P(1+r/m)^(tm)

20,000=P(1+(0.06)/1)^(3(1))=P(1.06)^3

(20,000)/((1.06)^3)=P

P=16,792.39

The accountant must deposit $ 16,792.39 .

(b) If only $ 15,000 is available to deposit now, what annual interest rate is required for it to increase to $ 20,000 in three years?

Here P = 15,000 , A = 20,000 , m = 1 , t= 3 , and r is unknown. Substitute the known values into the compound interest formula and solve for r .

20,000=15,000(1+r/1)^3

4/3=(1+r)^3 Divide both sides by 15,000

(4/3)^(1/3)=1+r Take the cubs root on both sides.

(4/3)^(1/3)-1=r Subtract 1 on both sides

r ≈ 0.10 Use a calculator.

An interest rate of 10 % will produce enough interest to increase the $ 15,000 deposit to the $ 20,000 needed at the end of three years.

Perhaps the single most useful exponential function is the function deﬁned by f(x) =e^x , where e is an irrational number that occurs often in practical applications. The number e comes up in a natural way when using the formula for Compound interest.

Suppose that a lucky investment produces an annual interest of 100 %, so that r=1.00 , or r=1 . Suppose also that only $ 1 1 can be deposited at this rate, and for only one year. Then P=1 and t=1 . Substitute into the formula for compound interest:

P(1+r/m)^(tm)=1(1+1/m)^(1(m))=(1+1/m)^m

As interest is compounded more and more often, the value of this expression will increase. If interest is compounded annually, making m = 1 , the total amount on deposit is

(1+1/m)^m=(1+1/1)^1=2^1=2 ,

so an investment of $ 1 becomes $ 2 in one year.

A calculator with a y^x key gives the results in the table at the left. These results have been rounded to ﬁve decimal places. The table suggests that, as m increases, the value of (1+1/m)^m gets closer and closer to some ﬁxed number. It turns out that this is indeed the case. This ﬁxed number is called e .

VALUE OF e To nine decimal places,

e ≈ 2.718281828

NOTE Values of e^x can be found with a calculator that has a key marked e^x or by using a combination of keys marked INV and ln x . See your instruction booklet for details or ask your instructor for assistance.

In Figure 5.6 the functions y=2^x , y=e^x , and y=3^x are graphed for comparison.

Figure 5.6

EXPONENTIAL GROWTH| AND DECAY As mentioned above, the number e is important as the base of an exponential function because many practical applications require an exponential function with base e . For example, it can be shown that in situations involving growth or decay of a quantity, the amount or number present at time t often can be closely approximated by a function deﬁned by

A=A_0e^(kt) ,

where A_0 is the amount or number present at time t = 0 and k is a constant.

The next example illustrates exponential growth.

SOLVING AN EXPONENTIAL GROWTH PROBLEM

The U. S. Consumer Price Index (CPI, or cost of living index) has risen exponentially over the years. From 1960 to 1990 , the CPI is approximated by

A(t)=34e^(0.04t)

where t is time in years, with t=0 corresponding to 1960 . The index in 1960, at t=0 , was

A(0)=34e^((0.04)(0))

= 34e^0

= 34 . e^0=1

To ﬁnd the CPI for 1990 , let t=1990-1960=30 , and find A(30) .

A(30)=34e^((0.04)(30))

= 34e^(1.2)

= 113 e^(1.2) ≈ 3.3201

The index measures the avenge change in prices relative to the base year 1983 ( 1983 corresponds to 100 ) of a common group of goods and services. Our result of 113 means that prices increased an average of 113-34=79 percent over the 30 -year period from 1960 to 1990 .

5.2 LOGARITHMIC FUNCTIONS

The previous section dealt with exponential functions of the form y=a^x for all positive values of a , where a!=1 . As mentioned there, the horizontal line test shows that exponential functions are one-to-one. and thus have inverse functions. In this section we discuss the inverses of exponential functions. The equation deﬁning the inverse of a function is found by exchanging x and y in the equation that deﬁnes the function. Doing so with y=a^x gives

a=a^y

as the equation of the inverse function of the exponential function deﬁned by y=a^x . This equation can be solved for y by using the following deﬁnition.

For all real numbers y , and all positive numbers a and x , where a!=1 :

y=log_a(x) if and only x=a^y .

The "log" in the deﬁnition above is an abbreviation for logarithm. Read log_a(x) as “the logarithm to the base a of x ." Intuitively, the logarithm to the base a of x is the power to which a must be raised to yield x .

In working with logarithms, it is helpful to remember the following.

MEANING OF log_a(x)

A logarithm is an exponent; log_a(x) is the exponent on the base a that yields the number x .

The “log” in y=log_a(x) is the notation for a particular function and there must be a replacement for x following it, as in log_a(3) , log_a(2x-1) , or log_a(x^2) . Avoid writing meaningless notation such as y=log or y=log_a .

CONVERTING BETWEEN EXPONENTIAL AND LOGARITHMIC STATEMENTS

The chart below shows several pairs of equivalent statements. The same statement is written in both exponential and logarithmic forms.

LOGARITHMIC EQUATIONS The deﬁnition of logarithm can be used to solve logarithmic equations, as shown in the next example.

SOLVING LOGARITHMIC EQUATIONS

Solve each equation.

(a) log_x(8/27)=3

First, write the expression in exponential form.

x^3=8/27

x^3=(2/3)^3 8/27=(2/3)^3

x=2/3 property (b) of exponents

The solution set is {2/3} .

(b) log_4(x)=5/2

In exponential form, the given statement becomes

4^(5/2)=x

(4^(1/2))^5=x

2^5=x

32=x

The solution set is {32} .

LOGARITHMIC FUNCTIONS The logarithmic function with base a is defined as follows.

LOGARITHMIC FUNCTIONS

If a>0 , a!=1 , and x>0 , then

f(x)=log_a(x)

defines the logarithmic function with base a .

Exponential and logarithmic functions are inverses of each other. Since the domain of an exponential function is the set of all real numbers. the range of a logarithmic function also will be the set of all real numbers. In the same way, both the range of an exponential function and the domain of a logarithmic function are the set of all positive real numbers, so logarithms can be found for positive numbers only.

The graph of y=2^x is shown in red in Figure 5.7. The graph of its inverse is found by reﬂecting the graph of y = 2^x about the line y = x . The graph of the inverse function. deﬁned by y=log_2(x) , shown in blue. has the y -axis as a vertical asymptote.

Figure 5.7

Figure 5.8

The graph of y=(1/2)^x is shown in red in Figure 5.8. The graph of its inverse, deﬁned by y=log_(1/2)x , in blue, is found by reﬂecting the graph of y = (1/2)^x about the line y = x . As Figure 5.8 suggests, the graph of y=log_(1/2)x also has the y-axis for a vertical asymptote.

The graphs of y = log_2x in Figure 5.7 and y=log_(1/2)x in Figure 5.8 suggest the following generalizations about the graphs of logarithmic functions of the form f(x) = log_a(x) .

GRAPH OF f(x)=log_a(x)

1. The point (1, 0) is on the graph.

3. The y -axis is a vertical asymptote.

4. The domain is (0,∞) and the range is (-∞,∞) .

Compare these generalizations to those for exponential functions discussed in Section 5.1.

More general logarithmic functions can be obtained by forming the composition of h(x) = log_a(x) with a function g(x) to get

f(x)=h[g(x)]=log_a[g(x) ]

The next examples illustrate some composite functions of this type.

GRAPHING A COMPOSITE LOGARITHMIC FUNCTION

Graph f(x)=log_2(x-1) .

The graph of this function will be the same as that of f(x) = log_2(x) , but shifted 1 unit to the right because x-1 is given instead of x . This makes the domain (1,∞) instead of (0,∞) . The line x = 1 is a vertical asymptote. The range is (-∞,∞) . Sec Figure 5.9.

Figure 5.9

GRAPHING A TRANSLATED LOGARITHMIC FUNCTION

Graph f(x)=(log_3x)-1 .

This function will have the same graph as that of g(x) = log_3x translated down 1 unit. A table of values is given below for both g(x) = log_3x and f(x)=(log_3x)-1 .

The graph is shown in Figure 5.10.

Figure 5.10

Graph y=log_3|x| .

Write y=log_3|x| in exponential form as 3^y=|x| to help identify some ordered pairs that satisfy the equation. (This is usually a good idea when graphing a logarithmic function.) Here. it is easier to choose y -values and ﬁnd the corresponding x -values. Doing so gives the following ordered pairs.

Plotting these points and connecting them with a Smooth curve gives the graph in Figure 5.11. The y -axis is a vertical asymptote. Notice that, since y=log_3|-x|=log_3|x| , the graph is symmetric with respect to the y -axis.

Figure 5.11

CAUTION If you write a logarithmic function in exponential form, choosing y -values to calculate x -values as we did in Example 5, be careful to get the ordered pairs in the correct order.

PROPERTIES OF LOGARITHMS Logarithms originally were important as an aid for numerical calculations, but the availability of inexpensive calculators has made this application of logarithms obsolete‘ Yet the principles behind the use of logarithms for calculation are important; these principles are based on the properties listed below.

PROPERTIES OF LOGARITHMS

For any positive real numbers x and y , real number r , and any positive real number a, a != 1 :

(a) log_a(xy)=log_a(x)+log_a(y)

(b) log_a(x/y)=log_a(x)-log_a(y)

(c) log_a(x^r)=rlog_a(x)

(d) log_a(a)=1

(e) log_a(1)=0

To prove Property (3), let

m=log_a(x) and n=log_a(y) .

By the deﬁnition of logarithm,

a^m=x and a^n=y .

Multiplication gives

a^m*a^n=xy .

By a Property of exponents,

a^(m+n)=xy .

Now use The deﬁnition of logarithm to write

log_a(xy)=m+n .

Since m=log_a(x) and n=log_a(y) .

log_a(xy)=log_a(x)+log_a(y) .

Properties (b) and (c) are proven in a similar way. (See Exercises 68 and 69.) Properties (d) and (e) follow directly from the deﬁnition of logarithm since a^1=a and a^0=1 . The properties of logarithms are useful for rewriting expressions with logarithms in different forms, as shown in the next examples.

USING THE PROPERTIES OF LOGARITHMS

Assuming that all variables represent positive real numbers, use the properties of logarithms to rewrite each of the following expressions.

(a) log_6(7*9)

log_6(7*9)=log_6(7)+log_6(9)

(b) log_9(15/7)

log_9(15/7)=log_9(15)-log_9(7)

(c) log_5root(8)

log_5root(8)=log_5(8^(1/2))=1/2log_5(8)

(d) log_a((mnq)/(p^2)) = log_a(m)+log_a(n)+log_a(q)-2log_a(p)

(e) log_(a)root(3,m^2)=2/3log_a(m)

(f) log_(b)root(n,(x^3y^5)/(z^m))

= (1/n)log_b((x^3y^5)/(z^m)

= 1/n(log_b(x^3)+log_b(y^5)-log_b(z^m)

= 1/n(3log_b(x)+5log_b(y)-mlog_b(z))

= (3/n)log_b(x)+(5/n)log_b(y)-(m/n)log_b(z)

Notice the use of parentheses in the second step. The factor 1/n applies to each term.

Use the properties of logarithms to write each of the following as a single logarithm with a coefficient of 1 . Assume that all variables represent positive real numbers.

(a) log_3(x+2)+log_3(x)-log_3(2)

Using Properties (a) and (b),

log_3(x+2)+log_3(x)-log_3(2)=log_3((x+2)x)/(2)

(b) 2log_a(m)-3log_a(n)=log_a(m^2)-log_a(n^3) = log_a(m^2/n^3)

Here we used Property (c), then Property (b).

(c) 1/2log_b(m)+3/2log_b(2n)-log_b(m^2n)

= log_b(m^(1/2))+log_b(2n)^(3/2)-log_b(m^2n) Property (c)

= log_b((m^(1/2)(2n)^(3/2))/(m^2n)) Properties (a) and (b)

= log_b((2^(3/2)n^(1/2))/(m^(3/2)) Rules for exponents

= log_b((2^3n)/(m^3))^(1/2) Rules for exponents

= log_b(root((8n)/(m^3)) Definition of a^(1/n)

There is no property of logarithms to rewrite a logarithm or’ a sum or difference. That is why, in Example 7(a), log_3 (x + 2) was not written as log_3x+log_3(2) Remember, log_3x+log_3(2)=log_3(x*2) .

USING THE PROPERTIES OF LOGARITHMS WITH NUMERICAL VALUES

Assume that log_10(2)=0.3010 . Find the base 10 logarithms of 4 and 5 . By the properties of logarithms,

log_10(4)=log_10(2^2)=2log_10(2)=2(0.3010)=0.6020

log_10(5)=log_10(10/2)=log_10(10)-log_10(2) = 1-0.3010=0.6990 .

We used Property (d) to replace log_10(10) with 1 .

Compositions of the exponential and logarithmic functions can be used to get two more useful properties. If f(x)=a^x and g(x)=log_a(x) , then

f[g(x)]=a^(log_a(x))

and g[f(x)]=log_a(a^x)

THEOREM ON INVERSES

For a>0 , a!=1 :

a^(log_a(x))=x and log_a(a^x)=x

PROOF

Exponential and logarithmic functions are inverses of each other, so f[g(x)]=x and g[f(x)]=x . Letting f(x)=a^x and g(x)=log_a(x) gives both results.

By the results of the last theorem,

log_5(5^3)=3 , 7^(log_7(10))=10 , and log_r(r^(k+1))=k+1 .

The second statement in the theorem will be useful in Sections 5.4 and 5.5 when solving logarithmic or exponential equations.

5.3 EVALUATING LOGARITHMS; CHANGE OF BASE

COMMON LOGARITHMS Base 10 logarithms are called common logarithms. The common logarithm of the number x , or log_10(x) . is often abbreviated as just log x , and we will use that convention from now on. A calculator with a log key can be used to ﬁnd base 10 logarithms of any positive number.

Example 1.

EVALUATING COMMON LOGARITHMS

Use a calculator to evaluate the following logarithms`.

(a) log 142

Enter 142 and press the log key. This may be a second function key on some calculators. With other calculators, these steps may be reversed. Consult your owner’s manual if you have any problem using this key. The result should be 2.152 to the nearest thousandth.

(b) log 0.005832

A calculator gives

log 0.005832 ≈ -2.234 .

NOTE Logarithms of numbers less than 1 are always negative, as suggested by the graphs in Section 5.2.

In chemistry, the pH of a solution is deﬁned as

pH=-log[H_3O^+0 ],

where [H_3O^+0 ] is the hydronium ion concentration in moles per liter. The pH value is a measure of the acidity or alkalinity of solutions. Pure water has a pH of 7.0 , substances with pH values greater than 7.0 are alkaline, and substances with pH values less than 7.0 are acidic.

(a) Find the pH of a solution with [H_3O^+0 ]= 2.5x10^-4

pH=-log[H_3O^+0 ]

pH=-log(2.5x10^-4) Substitute.

= -(log(2.5)+log10^-4) Property (B) of logarithms

= -(0.3979-4)

= -0.3979+4

= 3.6

It is customary to round pl-I values to the nearest tenth.

(b) Find the hydronium ion concentration of a solution with pH = 7.1 .

7.1=-log[H_3O^+0 ] Substitute.

-7.1=log[H_3O^+0 ] Multiply by -1 .

[H_3O^+0]=10^(-7.1) Write in exponential form.

Evaluate 10^(-7.1) with a calculator to get

[H_3O^+0 ] ≈ 7.9x10^-8 .

SOLVING AN APPLICATION OF BASE 10 LOGARITHMS

The loudness of sounds is measured in a uni! called a decibel. To measure with this unit, we ﬁrst assign an intensity of {Iota}_0 to a very faint sound, called the threshold sound. If a particular sound has intensity {Iota} , then the decibel rating of this louder sound is

d=10log {Iota}/{Iota}_0 .

Find the decibel rating of a sound with intensity 10,000{Iota}_0

d=10log (10,000{Iota}_0)/({Iota}_0)

= 10log10000

= 10(4) log10000=4

= 40

The sound has a decibel rating of 40 .

NATURAL LOGARITHMS In In most practical applications of logarithms, the number e ≈ 2.718281828 is used as base. The number e is irrational, like PI . Logarithms to base e are called natural logarithms, since they occur in the life sciences and economics in natural situations that involve growth and decay. The base e logarithm of x is written ln x (read “ el-en x "). A graph of the natural logarithm function deﬁned by f(x) = ln x is given in Figure 5.12. Natural logarithms can be found with a calculator that has an In key.

Figure 5.12

EVALUATING NATURAL LOGARITHMS

Use a calculator to ﬁnd the following logarithms.

(a) In 85

With a calculator, enter 85 , press the In key, and read the result, 4.4427 . The steps may be reversed with some calculators. If your calculator has an e^x key, but not a key labeled ln x , natural logarithms can be found by entering the number, pressing the {Iota}NV key and then the e^x key. This works because y=e^x is the inverse function of y=lnx (or y = log_e(x) .

(b) ln 127.8=4.850

(c) ln 0.049=-3.02

As with common logarithms, natural logarithms of numbers between 0 and 1 are negative.

APPLYING NATURAL LOGARITHMS

Geologists sometimes measure the age of rocks by using “atomic clocks." By measuring the amounts of potassium 40 and argon 40 in a rock, the age t of the specimen in years is found with the formula

t=(1.26x10^9)(ln[1+8.33(A/K)])/(ln2)

A and K are respectively the numbers of atoms of argon 40 and potassium 40 in the specimen.

(a) How old is a rock in which A = 0 and K > 0 ?

If A = 0 , A/K = 0 and the equation becomes

t=(1.26x10^9)(ln1)/(ln2)=(1.26x10^9)(0)=0

The ruck is 0 years old or new.

(b) The ratio A/K for a sample of granite from New Hampshire is 0.212 . How old is the sample?

Since A/K is 0.212 , we have

t=(1.26x10^9)(ln[1+8.33(0.212)])/(ln2)=1.85x10^9

The granite is about 1.85 billion years old.

LOGARITHMS TO OTHER BASES A calculator can be used to ﬁnd the values of either natural logarithms (base e ) or common logarithms (base 10 ). However, sometimes it is convenient to use logarithms to other bases. The following theorem can be used to convert logarithms from one base lo another.

CHANGE OF BASE THEOREM

For any positive real numbers x, a , and b . where a!=1 and b !=1 :

log_a(x)=(log_b(x))/(log_b(a)

This theorem is proved by using the deﬁnition of logarithm to write y = log_a(x) in exponential form.

Let y=log_a(x)

a^y=x Change to exponential form

log_b(a^y)=log_b(x) Take logarithms on both sides

ylog_b(a)=log_b(x) Property (c) of logarithms

y=(log_b(x))/(log_b(a)) Divide both sides by log_b(a) .

log_a(x)=(log_b(x))/(log_b(a Substitute log, x for y .

Any positive number other than 1 can be used for base b in the change of base rule, but usually the only practical bases are e and 10 , since calculators give logarithms only for these two bases.

NOTE Some calculators have only a log key or an ln key. In that case, the change of base rule can be used to ﬁnd logarithms to the missing base.

The next example shows how the change of base rule is used to ﬁnd logarithms to bases other than 10 or e with a calculator.

USING THE CHANGE OF BASE RULE

Use natural logarithms to ﬁnd each of the following. Round to the nearest hundredth.

(a) log_5(17)

Use natural logarithms and the change of base theorem.

log_5(17)=(log_e17)/(loge5)

= (ln17)/(ln5)

Now use a calculator to evaluate this quotient.

log_5(17) ≈ (2.8332)/(1.69094)

≈ 1.76

To check, use a calculator with a y‘ key, along with the deﬁnition of logarithm, to verify that 5^(1.76) ≈ 17 .

(b) log_2*1

log_2*1=(ln*1)/(ln*2) ≈ (-2.3026)/(0.6931)=-3.32

In Example 6, logarithms that were evaluated in the intermediate steps, such as ln 17 and ln 5 , were shown to four decimal places. However, the ﬁnal answers were obtained without rounding off these intermediate values, using all the digits obtained with the calculator. In general, it is best to wait until the ﬁnal step to round off the answer; otherwise, a build-up of round-off error may cause the ﬁnal answer to have an incorrect ﬁnal decimal place digit.

SOLVING AN APPLICATION WITH BASE 2 LOGARITHMS

One measure of the diversity of the species in an ecological community is given by the formula

H=-[P_1log_2O_1+P_2log_2P_2+...+P_(n)log_2P_n ],

where P_1,P_2,...,P_n are the proportions of a sample belonging to each of n species found in the sample. For example, in a community with two species, where there are 90 of one species and 10 of the other, P_1 = 90/100 = 0.9 and P_2 = 10/100 = 0.1 . Thus,

H=-[0.9log_2(0.9)+0.1log_2(0.1) ].

Example 6(b), log_2(0.1) was found to be -3.32 . Now ﬁnd log_2(0.9)

Therefore,

H ≈ -[(0.9)(0.152)+(0.1)(-3.32) ]≈ 0.469

If the number in each species is the same, the measure of diversity is 1 . representing “perfect” diversity. In a community with little diversity, H is close to 0 . In this example, since H = 0.5 . there is neither great nor little diversity.

5.4 EXPONENTIAL AND LOGARITHMIC EQUATIONS

As mentioned at the beginning of this chapter, exponential and logarithmic functions are important in many useful applications of mathematics. Using these functions in applications often requires solving exponential and logarithmic equations. Some simple equations were solved in the ﬁrst two sections of this chapter. More general methods for solving these equations depend on the properties below. These properties follow from the fact that exponential and logarithmic functions are one-to-one. Property 1 was given and used to solve exponential equations in Section 5.1.

PROPERTIES OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS

For b>0 and b!=1 :

1. b^x=b^y if and only if x=y .

2. x>0 and y>0 ,

log_b(x)=log_b(y) if and only if x=y .

EXPONENTIAL EQUATIONS The ﬁrst examples illustrate a general method, using Property 2, for solving exponential equations.

SOLVING AN EXPONENTIAL EQUATION

Solve the equation 7^x=12 .

In Section 5.1, we saw that Property 1 cannot be used to solve this equation, so we apply Property 2. While any appropriate base b can be used to apply Properly 2. the best practical base to use is base 10 or base e . Taking base e (natural) logarithms of both sides gives

7^x=12

ln7^x=ln12

xln7=ln12 Property (c) of logarithms

x=(ln12)/(ln7) Divide by ln 7 .

A decimal approximation for x can be found using a calculator:

x=(ln12)/(ln7) ≈ (2.4849)/(1.9459) ≈ 1.277 .

A calculator with a y^x key can be used to check this answer. Evaluate 7^(1.277) ; the result should be approximately 12 . This step veriﬁes that, to the nearest thousandth, the solution set is {1.277} .

Be careful when evaluating a quotient like (ln12)/(ln7) in Example 1. Do not confuse this quotient with ln(12/7) which can be written as ln 12 - ln 7 . You cannot change the quotient of two logarithm to a difference of logarithms.

(ln12)/(ln7)!=ln(12/7)

SOLVING AN EXPONENTIAL EQUATION

Solve e^(-2lnx)=1/16

Use a property of logarithms to rewrite the exponent on the left side of the equation.

e^(-2lnx)=1/16

e^(lnx^-2)=1/16 Property (0) of logarithms

x^-2=1/16 Theorem on inverses: e^(lnk)=k

x^-2=4^-2 1/16=1/4^2=4^-2

x=4 Property 1 given above

Check this answer by substituting in the original equation to see that the solution set is (4).

LOGARITHMIC EQUATIONS The next examples show some ways to solve logarithmic equations. The properties of logarithms given in Section 5.2 are useful here, as is Property 2.

SOLVING A LOGARITHMIC EQUATION

Solve log_a(x+6)-log_a(x+2)=log_a(x) .

Using a property of logarithms, rewrite the equation as

log_a(x+6)/(x+2)=log_a(x) Property (b) of logarithms

Now the equation is in the proper form to use Property 2.

(x+6)/(x+2)=x Property 2

(x+6)=x(x+2) Multiby by x+2 .

x+6=x^2+2x Distributive property

x^2+x-6=0 Get 0 on one side.

(x+3)(x-2)=0 Use the zero-factor property.

x=-3 or x=2 .

The negative solution (x = -3) cannot be used since it is not in the domain of log_a(x) in the original equation. For this reason, the only valid solution is the positive number 2 , giving the solution set {2}.

Recall that the domain of y=log_b(x) is (0,∞) . For this reason, it is always necessary to check that the solution of a logarithmic equation results in the logarithms of positive numbers in the original equation.

When physicians prescribe medication they must consider how the drug's effectiveness decreases over time. If, each hour, a drug is only 90 % as effective as the previous hour, at some point the patient will not be receiving enough medication and must receive another dose. This situation can be modeled with a geometric sequence (see Section 9.2). If the initial dose was 200 mg and the drug was administered 3 hours ago, the expression 200(0.90)^2 represents the amount of effective medication still available. Thus, 200(0.90)^2=162 mg are still in the system. To determine how long it would take for the medication to reach the dangerously low level of 50 mg, we consider the equation 200(0.90)^x=50 , which is solved using logarithms.

200(0.90)^x=50

(0.90)^x=0.25

log(0.90)^x=log(0.25)

xlog(0.90)=log(0.25)

x=(log(0.25))/(log(0.90)) ≈ 13.16

Since x represents n - 1 , the drug will reach a level of 50 mg in about 14 hours.

SOLVING LOGARITHMIC EQUATION

Solve log (3x+2)+log (x-1)=1

Since log x is an abbreviation for log_10(x) , and 1=log_10(10) , the properties of logarithms give

log (3x+2)(x-1)=log10

(3x+2)(x-1)=10 Property (a) of logarithms

3x^2-x-2=10 Property 2

3x^2-x-12=0

Now use the quadratic formula to get

x=(1+-root(1+144))/(6)

If x=((1-root(145))/(6) , then x-1<0 ; therefore. log (x - l) is not deﬁned and this proposed solution must be discarded, giving the solution set

{(1+root(145))/(6)} .

The deﬁnition of logarithm could have been used in Example 4 by ﬁrst writing

log (3x+2) + log (x-1)=1

log_10(3x+2)(x-1)=1 Property (a)

(3x+2)(x-1)=10^1 Definition of logarithm

then continuing as shown above.

Solve ln e^(lnx)-ln(x-3)=ln2

On the left, ln e^(lnx) can be written as ln x using the theorem on inverses at the end of Section 5.2. The equation becomes

lnx-ln(x-3)=ln2

ln(x/(x-3))=ln2 Property (b)

x/(x-3)=2 Property 2

x=2x-6 Multiply by x-3 .

6=x .

Verify that the solution set is {6} .

A summary of the methods used for solving equations in this section follows.

SOLVING EXPONENTIAL AND LOGARITHMIC EQUATIONS

An exponential or logarithmic equation may be solved by changing the equation into one of the following forms, where a and b are real numbers, a > 0 , and a!=1 .

1. a^(f(x))=b

Solve by taking logarithms of each side. (Natural logarithms are often a good choice.)

2. log_(a)f(x)=log_(a)g(x)

From the given equation, f(x) = g(x) , which is solved algebraically.

3. log_(a)f(x) = b

Solve by using the deﬁnition of logarithm to write the expression in exponential form as f(x) = a^b .

The next examples show applications of exponential and logarithmic equations.

SOLVING A COMPOSITE EXPONENTIAL EQUATION

The strength of a habit is a function of the number of times the habit is repeated. If N is the number of repetitions and H is the strength of the habit, then, according to psychologist C. L. Hull,

(H)/(1000)=1-e^(-kN) Divide by 1000 .

(H)/(1000)-1=-e^(-kN) Subtract 1 .

e^(-kN)=1-(H)/(1000) Multiply by -1 .

Now solve for k . As shown earlier, we take logarithms on each side of the equation and use the fact that ln e^x = x .

lne^(-kN)=ln(1-(H)/(1000))

-kN=ln(1-(H)/(1000)) lne^x=x

k=-1/(N)ln(1-(H)/(1000)) Multiply by -1/(N) .

With the last equation, if one pair of values for H and N is known, k can be found, and the equation can then be used to ﬁnd either H or N , for given values of the other variable.

SOLVING COMPOSITE LOGARITHMIC EQUATION

In the exercises for Section 5.3, we saw that the number of species in a sample is given by S , where

S=aln(1+n/a) ,

n is the number of individuals in the sample, and a is a constant. Solve this equation for n .

We begin by solving for (1+n/a) Then we can change to exponential form and solve the resulting equation for n .

S/a=ln(1+n/a) Divide by a .

e^(S/a)-1=n/a Write in exponential form

e^(S/a)-1=n/a Subtract 1 .

n=a(e^(S/a)-1)) Multiply by a .

Using this equation and given values of S and a , the number of species in a sample can be found.

5.5 EXPONENTIAL GROWTH AND DECAY

In many cases, quantities grow or decay according to a function deﬁned by

A(t)=A_0e^(kt) .

As mentioned in Section 5.1, when k is positive, the result is a growth function; when k is negative, it is a decay function. This section gives several examples of applications of this function.

CONTINUOUS COMPOUNDING The compound interest formula

was discussed in Section 5.1. The table presented there shows that increasing the frequency of compounding makes smaller and smaller differences in the amount of interest earned. In fact, it can be shown that even if interest is compounded at intervals of time as small as one chooses (such as each hour, each minute. or each second), the total amount of interest earned will be only slightly more than for daily compounding. This is true even for a process called continuous compounding. As suggested in Section 5.1, the value of the expression (1+1/m)^m approaches e as m gets larger. Because of this, the formula for continuous compounding involves the number e .

CONTINUOUS COMPOUNDING If P dollars is deposited at a rate of interest r compounded continuously for t years, the final amount on deposit is

A=Pe^(rt)

dollars.

SOLVING A CONTINUOUS COMPOUNDING PROBLEM

Suppose $ 5000 is deposited in an account paying 8 % compounded continuously for ﬁve years. Find the total amount on deposit at the end of ﬁve years.

Let P=5000 , t=5 , and r=0.08 / Then

A=5000e^(0.08(5))=5000e^4

Using a calculator. we ﬁnd that e^(0.4) ≈ 1.49182 , and

A=5000e^(0.4)=7459.12 ,

or $ 7459.12 . As a comparison, the compound interest formula with daily compounding gives

= 5000(1+(0.08)/365)^(5(365))=7458.80 ,

about 32 ¢ less.

SOLVING CONTINUOUS COMPOUNDING PROBLEM

How long will it take for the money in an account that is compounded continuously at 8% interest to double?

Use the formula for continuous compounding, A=Pe^(rt) , to ﬁnd the time t that makes A=2P Substitute 2P for A and 0.08 for r , then solve for t .

2P=Pe^(0.08t) Substitute.

2=e^(0.08t) Divide by P .

Taking natural logarithms on both sides gives

ln2=lne^(0.08t) .

Use the property lne^x=x to get lne^(0.08t)=0.08t .

ln2=0.08t

(ln2)/(0.08)=t Substitute.

8.664=t Divide by 0.08 .

It will take about 8 2/3 years for the amount to double.

GROWTH AND DECAY The next examples illustrate applications of exponential growth and decay.

SOLVING AN EXPONENTIAL DECAY PROBLEM

Nuclear energy derived from radioactive isotopes can be used to supply power to space vehicles. The output of the radioactive power supply for a certain satellite is given by the function

y=40e^(-0.004t) ,

where y is in watts and t is the time in days.

(a) How much power will be available at the end of 180 days?

Let t=180 in the formula.

y=40e^(-0.004(180))

y ≈ 19.5 Use a calculator.

(b) How long will it take for the amount of power to be half of its original strength?

The original amount of power is 40 watts. (Why?) Since half of 40 is 20 , replace y with 20 in the formula, and solve for t .

20=40e^(-0.004t)

0.5=e^(-0.004t) Divide by 40

ln 0.5=lne^(-0.004t)

ln 0.5=-0.004t lne^x=x

t=(ln(1/2))/(-0.004)

t ≈ 173 Use a calculator.

After about 173 days, the amount of available power will be half of its original amount.

In Examples 2 and 3(b), we found the amount of time that it would take for an amount to double and to become half of its original amount. These are examples of doubling time and half-life. The doubling time of a quantity that grows exponentially is the amount of time that it takes for any initial amount to grow to twice its value. Similarly, the half-life of a quantity that decays exponentially is the amount of time that it takes for any initial amount to decay to half its value.

SOLVING AN EXPONENTIAL

Carbon 14 is a radioactive form of carbon that is found in all living plants and animals. After a plant or animal dies, the radiocarbon disintegrates. Scientists determine the age of the remains by comparing the amount of carbon 14 present with the amount found in living plants and animals. The amount of carbon 14 present after t years is given by the exponential equation

A(t)=A_0e^(kt)

With k ≈ -(ln2)(1/5700) .

(a) Find the half-life.

Let A(t)=(1/2)A_0 and k=-(ln2)(1/5700) .

1/2A_0=A_0e^(-(ln2)(1/57000t)

1/2=e^(-(ln2)(1/5700)t) Divide by A_0 .

ln(1/2)=lne^(-(ln2)(1/5700)t) Take logarithms on both sides.

ln(1/2)=-(ln2)/(5700)t lne^x=x

-5700/(ln2)ln(1/2)=t Multiply by -5700/(ln2)

-5700/(ln2)(ln1-ln2)=t Property (b)

-5700/(ln2)(-ln2)=t ln1=0

5700=t

The half-life is 5700 years.

(b) Charcoal from an ancient ﬁre pit on Java contained 1/4 the carbon 14 of a living sample of the same size. Estimate the age of the charcoal.

Let A(t)=1/4A_0 and k=-(ln2)(1/5700) .

1/4A_0=A_0e^(-(ln2)(1/5700)t)

1/4=e^(-(ln2)(1/5700)t)

ln(1/4)=lne^(-(ln2)(1/5700)t)

ln(1/4)=(ln2)/(5700)t

-5700/(ln2)ln(1/4)=t

t=11400

The charcoal is about 11400 years old.

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- Mastering Exponential Equations: A Step-by-Step Guide to Solving

Welcome to Warren Institute! In this article, we will explore the fascinating world of exponential equations and learn how to solve them effectively. Exponential equations, with their exponential functions, play a crucial role in various fields, such as finance, science, and engineering. Understanding how to solve these equations is essential for any aspiring mathematician or problem solver. Through clear explanations and step-by-step examples, we will delve into the techniques and strategies to tackle exponential equations confidently. Join us on this mathematical journey and discover the power of exponential equations!

## Understanding Exponential Equations

Basic properties of exponents, solving linear exponential equations, applications of exponential equations, how do you solve exponential equations with the same base on both sides, what are the steps to solve exponential equations with different bases, can you explain the concept of logarithms and how they are used to solve exponential equations, are there any special techniques or strategies for solving exponential equations involving fractions or negative exponents, is there a general approach or method for solving exponential equations that can be applied to different types of problems.

Exponential equations are mathematical expressions that involve variables raised to a power. It is important to have a clear understanding of how exponential equations work in order to solve them effectively. This section will provide an overview of the key concepts and strategies involved in solving exponential equations.

Understanding the basic properties of exponents is crucial when solving exponential equations. This section will cover the rules for multiplying, dividing, raising to a power, and taking the root of exponential expressions. By applying these properties correctly, you can simplify complex equations and make them easier to solve.

Solving linear exponential equations involves finding the value or values of the variable that make the equation true. This section will demonstrate step-by-step methods for solving equations with a single exponential term, as well as equations with multiple exponential terms. It will also cover different techniques, such as using logarithms, to solve more challenging exponential equations.

Exponential equations have various applications in real-world scenarios, such as population growth, compound interest, and radioactive decay. This section will explore how exponential equations can be used to model and solve problems in different fields, highlighting the practical significance of understanding and solving these equations in a mathematics education context.

## frequently asked questions

To solve exponential equations with the same base on both sides, you can use the property of equality for exponents. First, set the exponents equal to each other: (a^x = a^y) Next, equate the corresponding powers: (x = y) Finally, solve for the unknown variable.

The steps to solve exponential equations with different bases are:

1. Identify the different bases in the equation. 2. Use logarithms to write each base in terms of a common base. 3. Apply the properties of logarithms to simplify the equation. 4. Solve the resulting linear or quadratic equation for the variable. 5. Check the solution obtained by substituting it back into the original equation. 6. If necessary, state any restrictions on the domain of the variables.

Logarithms are mathematical functions that represent the inverse operation of exponentiation. They are used to solve exponential equations by converting them into logarithmic form, which makes it easier to manipulate and find the unknown variable. Logarithms help us to solve for the exponent (unknown variable) by finding the power to which a base must be raised to obtain a given number. The formula for solving an exponential equation using logarithms is: log(base b)(x) = y . In this equation, b represents the base, x is the number we want to find the exponent for, and y is the given value. By taking the logarithm of both sides of the equation, we can isolate the exponent and solve for it.

Yes, there are specific techniques and strategies for solving exponential equations involving fractions or negative exponents. One common approach is to use the properties of exponents, such as the power rule and the quotient rule, to simplify the equation and isolate the base. Another strategy is to convert the negative exponent into a positive exponent by taking the reciprocal of the base. Additionally, when dealing with fractional exponents, we can use the property that a fractional exponent represents a root.

Yes, there is a general approach for solving exponential equations that can be applied to different types of problems. The first step is to rewrite the equation so that both sides have the same base. Then, we can equate the exponents and solve for the variable by using logarithms.

In conclusion, understanding how to solve exponential equations is a crucial skill for students in mathematics education. By utilizing fundamental principles such as the laws of exponents, logarithms, and algebraic techniques, students can successfully navigate these challenging equations. Remembering the importance of accuracy and precision while applying appropriate methods is key. Furthermore, practicing with various examples and seeking guidance from educators or online resources can further enhance proficiency in this area. With a solid foundation in solving exponential equations, students will be well-equipped to tackle more advanced mathematical concepts and challenges in their academic journey.

If you want to know other articles similar to Mastering Exponential Equations: A Step-by-Step Guide to Solving you can visit the category Algebra .

Michaell Miller

Michael Miller is a passionate blog writer and advanced mathematics teacher with a deep understanding of mathematical physics. With years of teaching experience, Michael combines his love of mathematics with an exceptional ability to communicate complex concepts in an accessible way. His blog posts offer a unique and enriching perspective on mathematical and physical topics, making learning fascinating and understandable for all.

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## How to Graph an Exponential Function – A Step-by-Step Guide

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## Creating a Table of Values

Plotting points on the graph, identifying asymptotes and end behavior, transforming exponential graphs, real-world exponential models.

To graph an exponential function , I start by identifying the function’s base, which determines whether the function represents exponential growth or exponential decay .

For example, a function like $y = 5^x$ exhibits growth since the base is greater than 1. I plot points by choosing values for x and calculating the corresponding y values to sketch the curve of the function on a graph.

Exponential functions are vital in various real-world applications, including finance , where they model compound interest, life sciences for population growth, computer science for algorithmic complexity, and forensics for decay processes.

In each case, understanding the direction and shape of the graph helps in predicting future values. If the base of the function is less than 1, then the graph shows decay, representing processes like radioactive decay or depreciation.

To create an accurate representation, I ensure to plot key points such as the y-intercept at $y = b^0$, where b is the base, which always equals 1.

This sets the groundwork for recognizing an exponential function’s signature curve, with one side approaching zero—the horizontal asymptote—and the other rising or falling sharply depending on whether the function models growth or decay. Stay tuned as I show how this dynamic function can bring to light fascinating trends and predictions.

## Setting Up the Graph of an Exponential Function

When I’m about to graph an exponential function , my focus is on the behavior of the function, domain and range values, and the graphical characteristics like the horizontal asymptote.

Here’s how I set up the graph of an exponential function, step-by-step.

Firstly, I like to create a table of values to organize my x and y data points. For a function like $y = b^x$, where b is the base , I choose several x values — positive, negative, and zero. Then I calculate the corresponding y values, which show me the function’s growth or decay .

Table of Values Example:

Using the table of values , I plot each pair of x and y coordinates on the graph. This gives me a series of points that, when connected, start to show the characteristic shape of the exponential curve .

I also need to determine the horizontal asymptote of the exponential function. For most exponential functions , the horizontal asymptote is the line ( y = 0 ). It’s where the graph approaches but never touches as ( x ) progresses towards negative or positive infinity.

This ties into the end behavior of the function; for a growth function, ( y ) will increase without bound as ( x ) goes to infinity. For decay , the function approaches the horizontal asymptote as ( x ) increases.

When I graph the exponential functions , I often use various transformations of functions to modify the shape and position of the graph. Understanding these can help anyone tailor the function to fit their data or situation better.

The basic form of an exponential function is $f(x) = b^x$, where $b$ is a positive constant. From this parent function, I can apply different transformations to shift, reflect, stretch, or compress the graph.

Vertical shifts are achieved by adding or subtracting a constant value, which moves the graph up or down. For instance, $f(x) = b^x + k$ would result in a shift up if $k > 0$ and a shift down if $k < 0$.

Horizontal shifts involve adding or subtracting a constant to the input variable, $x$. The function $f(x) = b^{(x-h)}$ represents a shift to the right by $h$ units if $h > 0$ and to the left if $h < 0$.

Reflections are another vital transformation that flips the graph over a specific axis. To reflect a graph across the x-axis, I multiply the function by -1, yielding $f(x) = -b^x$. For a reflection across the y-axis, I would change the sign of the exponent, giving me $f(x) = b^{-x}$.

graphing exponential functions 3 1

Lastly, stretches and compressions change the graph’s steepness or width. Multiplying the function by a value $a > 1$ will stretch it vertically, while $0 < a < 1$ will compress it. Similarly, multiplying the exponent by $c > 1$ compresses the graph horizontally, and if $0 < c < 1$, it stretches it.

Here is a quick reference table for these transformations:

By applying these transformations systematically, I can shape the graph of an exponential function to fit the data or convey specific information visually. With some practice, these transformations become intuitive, and I can quickly sketch the modified graphs of exponential functions.

In my daily encounters, I often come across real-world applications of exponential functions that pique my curiosity. In finance , for instance, the concept is at the heart of understanding compound interest.

Here, the equation to calculate compound interest is $A = P(1 + \frac{r}{n})^{nt}$, where (A) is the amount of money accumulated after (n) years, including interest, (P) is the principal amount, (r) is the annual interest rate, (n) is the number of times that interest is compounded per year, and (t) is the time the money is invested for.

When I turn to forensics , something interesting happens. Exponential equations help me determine the time of death by analyzing the rate of cooling of the human body, which is an application of Newton’s Law of Cooling: $T(t) = T_e + (T_b – T_e)e^{-kt}$.

In computer science , algorithms for certain types of problems, such as sorting algorithms or calculating Fibonacci numbers, can have exponential growth in their runtime, depending on the algorithm’s complexity. This is crucial for making predictions about performance and efficiency.

The life sciences frequently utilize exponential graphs to model population growth or the spread of diseases, which can be dictated by equations like $P(t) = P_0e^{rt}$, where (P(t)) is the population at the time (t), (P_0) is the initial population, (r) is the growth rate, and (t) is time.

Each exponential model starts with a parent function of the form $f(x) = b^x$, which can be tailored to fit specific data and scenarios in the real world. Through analyzing exponential graphs , we can visually grasp the immense impact of exponential growth or decay and better understand the phenomena at hand.

I find that recognizing the power and implications of exponential functions in these areas can be quite insightful and empowering.

I hope my guide on graphing exponential functions has been informative and easy to grasp. When plotting exponential functions such as $y = b^x $, remember to identify a few crucial points. Ideally, you should include the y-intercept $(x=0), (y=1)$ and another point like (x=1), (y=b)) to ensure accuracy in your graph.

Always draw the horizontal asymptote, typically the line (y=0), and recognize its importance in showing the boundary that the function value will never reach.

A smooth curve that connects your plotted points will help visualize the rapid increase or decrease of the function.

Through practice and patience, you’ll find graphing these functions to be straightforward. Keep in mind that the base (b) greatly influences the shape of the curve: if (b > 1), you’re looking at growth, and if (0 < b < 1), it’s a decay.

Understanding how to graph exponential functions is not just about plotting points; it’s about recognizing patterns and behaviors that are fundamental to many natural processes and financial models.

So, next time you see a exponential curve , you’ll appreciate the mathematical beauty and the real-world implications it represents. Keep exploring, and don’t hesitate to revisit the sections on Creating a Table of Points or Understanding Exponential Growth for a refresher.

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Steps to Solve There are different kinds of exponential equations. We will focus on exponential equations that have a single term on both sides. These equations can be classified into 2 types. Type #1: Same Bases like : 4x = 49 4 x = 4 9 . Type #2: Different Bases like: 43 = 2x 4 3 = 2 x . (1 4)x = 32 ( 1 4) x = 32 (Part II below)

Method 1 Equating Two Exponents with the Same Base Download Article 1 Determine whether the two exponents have the same base. The base is the big number in an exponential expression. [1] You can only use this method when you are presented with an equation that has an exponent on either side, and each exponent has the same base.

Thus, we find the base b by dividing the y value of any point by the y value of the point that is 1 less in the x direction which shows an exponential growth. It works the same for decay with points (-3,8). (-2,4) (-1,2) (0,1), So 1/2=2/4=4/8=1/2. It will also have a asymptote at y=0. Next, if we have to deal with a scale factor a, the y ...

There are two methods for solving exponential equations. One method is fairly simple but requires a very special form of the exponential equation. The other will work on more complicated exponential equations but can be a little messy at times. Let's start off by looking at the simpler method.

21K 1.7M views 6 years ago New Algebra Playlist This algebra video tutorial explains how to solve exponential equations using basic properties of logarithms and natural logs. Examples include...

How to solve an exponential equation. For more in-depth math help check out my catalog of courses. Every course includes over 275 videos of easy to follow an...

To solve an exponential equation start by isolating the exponential expression on one side of the equation. Then, take the logarithm of both sides of the equation to convert the exponential equation into a logarithmic equation. The logarithm must have the same base as the exponential expression in the equation.

\ (x log 7=log 3\) Now, dividing both sides by \ (log 7\): \ (x=\frac {log 3} {log 7}\) Exponential Equations - Example 2: Solve the equation \ (4^ {2x-1}=64\). The bases are not the same, but we can rewrite \ (64\) as a base of \ (4\) → \ (4^3=64\)

One problem is is in step 4, you cannot factor out things using multiplication, factoring out is used with adding terms. Lets use simpler terms (2^2*2^3)^3*2^2 = (4*8)^3*4 = 131072. However, 2^2(2^3)^3=2048, so these two are clearly not the same. ... Let's get some practice solving some exponential equations, and we have one right over here. We ...

Quiz Course How to Solve Exponential Equations? When solving exponential equations, it is important to note what the base number is of the equation.

Step 1: Identify the base and the exponents. Base: 3 Exponent 1: 4 Exponent 2: 2 Step 2: Apply the product of powers property. 3^4 * 3^2 = 3^ (4 + 2) = 3^6 = 729 Property 2: Quotient of Powers a^m / a^n = a^ (m - n) This property states that when you divide two powers with the same base, you can subtract their exponents.

Help Tutorial Solve an equation, inequality or a system. Example: 2x-1=y,2y+3=x To see your tutorial, please scroll down Math Articles Exponential function Exponential and logarithmic function 5.1 EXPONENTIAL FUNCTIONS Recall from Chapter 1 the deﬁnition of ar, where r is a rational number: if r=mn, then for appropriate values of m and n,

Let's start off this section with the definition of an exponential function. If b b is any number such that b > 0 b > 0 and b ≠ 1 b ≠ 1 then an exponential function is a function in the form, f (x) = bx f ( x) = b x. where b b is called the base and x x can be any real number. Notice that the x x is now in the exponent and the base is a ...

Solving exponential equations with logarithms Solving exponential equations using logarithms Google Classroom Learn how to solve any exponential equation of the form a⋅b^ (cx)=d. For example, solve 6⋅10^ (2x)=48. The key to solving exponential equations lies in logarithms! Let's take a closer look by working through some examples.

😱The unknown is in the exponent!?? I'll show you step-by-step how to solve. ️Watch for growth and decay problems - 💕It really helps me here on YouTube w...

By applying these properties correctly, you can simplify complex equations and make them easier to solve. Solving Linear Exponential Equations. Solving linear exponential equations involves finding the value or values of the variable that make the equation true. This section will demonstrate step-by-step methods for solving equations with a ...

Step 1: Isolate the exponential and then apply the logarithm to both sides. Step 2: Apply the power rule for logarithms and write the exponent as a factor of the base. Step 3: Solve the resulting equation.

Symbolab is the best step by step calculator for a wide range of math problems, from basic arithmetic to advanced calculus and linear algebra. It shows you the solution, graph, detailed steps and explanations for each problem. Is there a step by step calculator for physics?

This video by Fort Bend Tutoring shows the process of solving exponential equations using logarithmic properties, natural logarithmic properties, substitutio...

The basic form of an exponential function is f ( x) = b x, where b is a positive constant.

Learn how to solve exponential equations using the Decimal Approach and the Standard Approach. Quick and easy explanation by PreMath.com1) 2^(2x-3) = 5^(1-x)...

476 likes, 2 comments - maths.visualization on February 13, 2024: "Logarithmic differentiation is a technique used to differentiate functions that are not easily di..." MATH. V I Z on Instagram: "Logarithmic differentiation is a technique used to differentiate functions that are not easily differentiable using basic differentiation rules.

Learn how to solve exponential equations with different bases using logarithms. Step-by-Step Technique by PreMath.com

In this video, we're going to teach you how to solve system of equations!Systems of equations are a common type of equation that you'll see on the AP Calculu...