Skip to content Skip to navigation

Portal

You are here: Home » Content » Exponential and Logarithmic Equations

Navigation

Recently Viewed

This feature requires Javascript to be enabled.
 

Exponential and Logarithmic Equations

Module by: First Last. E-mail the author

Summary: In this section, you will:

  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.

Note: You are viewing an old style version of this document. The new style version is available here.

Figure 1: Wild rabbits in Australia. The rabbit population grew so quickly in Australia that the event became known as the “rabbit plague.” (credit: Richard Taylor, Flickr)
Seven rabbits in front of a brick building.

In 1859, an Australian landowner named Thomas Austin released 24 rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers b, b, S, S, and T, T, where b>0, b1, b>0, b1, b S = b T b S = b T if and only if S=T. S=T.

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation 3 4x7 = 3 2x 3 . 3 4x7 = 3 2x 3 . To solve for x, x, we use the division property of exponents to rewrite the right side so that both sides have the common base, 3. 3. Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for x: x:

3 4x7 = 3 2x 3 3 4x7 = 3 2x 3 1 Rewrite 3 as 3 1 . 3 4x7 = 3 2x1 Use the division property of exponents. 4x7 =2x1  Apply the one-to-one property of exponents. 2x =6 Subtract 2x and add 7 to both sides. x =3 Divide by 3. 3 4x7 = 3 2x 3 3 4x7 = 3 2x 3 1 Rewrite 3 as 3 1 . 3 4x7 = 3 2x1 Use the division property of exponents. 4x7 =2x1  Apply the one-to-one property of exponents. 2x =6 Subtract 2x and add 7 to both sides. x =3 Divide by 3.

A General Note: Using the One-to-One Property of Exponential Functions to Solve Exponential Equations:

For any algebraic expressions S and T, S and T, and any positive real number b1, b1,

b S = b T if and only ifS=T b S = b T if and only ifS=T
(2)

How To:

Given an exponential equation with the form b S = b T , b S = b T , where S S and T T are algebraic expressions with an unknown, solve for the unknown.

  1. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S = b T . b S = b T .
  2. Use the one-to-one property to set the exponents equal.
  3. Solve the resulting equation, S=T, S=T, for the unknown.

Example 1

Problem 1

Solving an Exponential Equation with a Common Base

Solve 2 x1 = 2 2x4 . 2 x1 = 2 2x4 .

Solution
  2 x1 = 2 2x4 The common base is  2.     x1=2x4 By the one-to-one property the exponents must be equal.          x=3 Solve for x.   2 x1 = 2 2x4 The common base is  2.     x1=2x4 By the one-to-one property the exponents must be equal.          x=3 Solve for x.

Try It:

Exercise 1

Solve 5 2x = 5 3x+2 . 5 2x = 5 3x+2 .

Solution

x=2 x=2

Rewriting Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation 256= 4 x5 . 256= 4 x5 . We can rewrite both sides of this equation as a power of 2. 2. Then we apply the rules of exponents, along with the one-to-one property, to solve for x: x:

256= 4 x5 2 8 = ( 2 2 ) x5 Rewrite each side as a power with base 2. 2 8 = 2 2x10 Use the one-to-one property of exponents. 8=2x10 Apply the one-to-one property of exponents. 18=2x Add 10 to both sides. x=9 Divide by 2. 256= 4 x5 2 8 = ( 2 2 ) x5 Rewrite each side as a power with base 2. 2 8 = 2 2x10 Use the one-to-one property of exponents. 8=2x10 Apply the one-to-one property of exponents. 18=2x Add 10 to both sides. x=9 Divide by 2.

How To:

Given an exponential equation with unlike bases, use the one-to-one property to solve it.

  1. Rewrite each side in the equation as a power with a common base.
  2. Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form b S = b T . b S = b T .
  3. Use the one-to-one property to set the exponents equal.
  4. Solve the resulting equation, S=T, S=T, for the unknown.

Example 2

Problem 1
Solving Equations by Rewriting Them to Have a Common Base

Solve 8 x+2 = 16 x+1 . 8 x+2 = 16 x+1 .

Solution
     8 x+2 = 16 x+1 ( 2 3 ) x+2 = ( 2 4 ) x+1 Write  8 and 16 as powers of  2.     2 3x+6 = 2 4x+4 To take a power of a power, multiply exponents.    3x+6=4x+4 Use the one-to-one property to set the exponents equal.            x=2 Solve for x.      8 x+2 = 16 x+1 ( 2 3 ) x+2 = ( 2 4 ) x+1 Write  8 and 16 as powers of  2.     2 3x+6 = 2 4x+4 To take a power of a power, multiply exponents.    3x+6=4x+4 Use the one-to-one property to set the exponents equal.            x=2 Solve for x.

Try It:

Exercise 2

Solve 5 2x = 25 3x+2 . 5 2x = 25 3x+2 .

Solution

x=1 x=1

Example 3

Problem 1
Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve 2 5x = 2 . 2 5x = 2 .

Solution
2 5x = 2 1 2 Write the square root of  2 as a power of  2. 5x= 1 2 Use the one-to-one property. x= 1 10 Solve for x. 2 5x = 2 1 2 Write the square root of  2 as a power of  2. 5x= 1 2 Use the one-to-one property. x= 1 10 Solve for x.

Try It:

Exercise 3

Solve 5 x = 5 . 5 x = 5 .

Solution

x= 1 2 x= 1 2

Q&A:

Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Example 4

Problem 1
Solving an Equation with Positive and Negative Powers

Solve 3 x+1 =−2. 3 x+1 =−2.

Solution

This equation has no solution. There is no real value of x x that will make the equation a true statement because any power of a positive number is positive.

Analysis

Figure 2 shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Figure 2
Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.

Try It:

Exercise 4

Solve 2 x =−100. 2 x =−100.

Solution

The equation has no solution.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since log( a )=log( b ) log( a )=log( b ) is equivalent to a=b, a=b, we may apply logarithms with the same base on both sides of an exponential equation.

How To:

Given an exponential equation in which a common base cannot be found, solve for the unknown.

  1. Apply the logarithm of both sides of the equation.
    • If one of the terms in the equation has base 10, use the common logarithm.
    • If none of the terms in the equation has base 10, use the natural logarithm.
  2. Use the rules of logarithms to solve for the unknown.

Example 5

Problem 1

Solving an Equation Containing Powers of Different Bases

Solve 5 x+2 = 4 x . 5 x+2 = 4 x .

Solution
           5 x+2 = 4 x There is no easy way to get the powers to have the same base.         ln 5 x+2 =ln 4 x Take ln of both sides.     (x+2)ln5=xln4 Use laws of logs.   xln5+2ln5=xln4 Use the distributive law.   xln5xln4=2ln5 Get terms containing x on one side, terms without x on the other. x(ln5ln4)=2ln5 On the left hand side, factor out an x.           xln( 5 4 )=ln( 1 25 ) Use the laws of logs.                   x= ln( 1 25 ) ln( 5 4 ) Divide by the coefficient of x.            5 x+2 = 4 x There is no easy way to get the powers to have the same base.         ln 5 x+2 =ln 4 x Take ln of both sides.     (x+2)ln5=xln4 Use laws of logs.   xln5+2ln5=xln4 Use the distributive law.   xln5xln4=2ln5 Get terms containing x on one side, terms without x on the other. x(ln5ln4)=2ln5 On the left hand side, factor out an x.           xln( 5 4 )=ln( 1 25 ) Use the laws of logs.                   x= ln( 1 25 ) ln( 5 4 ) Divide by the coefficient of x.

Try It:

Exercise 5

Solve 2 x = 3 x+1 . 2 x = 3 x+1 .

Solution

x= ln3 ln( 2 3 ) x= ln3 ln( 2 3 )

Q&A:

Is there any way to solve 2 x = 3 x ? 2 x = 3 x ?

Yes. The solution is 0. 0.

Equations Containing e

One common type of exponential equations are those with base e. e. This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base e e on either side, we can use the natural logarithm to solve it.

How To:

Given an equation of the form y=A e kt , y=A e kt , solve for t. t.

  1. Divide both sides of the equation by A. A.
  2. Apply the natural logarithm of both sides of the equation.
  3. Divide both sides of the equation by k. k.

Example 6

Problem 1
Solve an Equation of the Form y = Aekt

Solve 100=20 e 2t . 100=20 e 2t .

Solution
100 =20 e 2t 5 = e 2t Divide by the coefficient of the power. ln5 =2t Take ln of both sides. Use the fact that ln(x) and  e x  are inverse functions. t = ln5 2 Divide by the coefficient of t. 100 =20 e 2t 5 = e 2t Divide by the coefficient of the power. ln5 =2t Take ln of both sides. Use the fact that ln(x) and  e x  are inverse functions. t = ln5 2 Divide by the coefficient of t.
Analysis

Using laws of logs, we can also write this answer in the form t=ln 5 . t=ln 5 . If we want a decimal approximation of the answer, we use a calculator.

Try It:

Exercise 6

Solve 3 e 0.5t =11. 3 e 0.5t =11.

Solution

t=2ln( 11 3 ) t=2ln( 11 3 ) or ln ( 11 3 ) 2 ln ( 11 3 ) 2

Q&A:

Does every equation of the form y=A e kt y=A e kt have a solution?

No. There is a solution when k0, k0, and when y y and A A are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is 2=−3 e t . 2=−3 e t .

Example 7

Problem 1
Solving an Equation That Can Be Simplified to the Form y = Aekt

Solve 4 e 2x +5=12. 4 e 2x +5=12.

Solution
4 e 2x +5=12 4 e 2x =7 Combine like terms. e 2x = 7 4 Divide by the coefficient of the power. 2x=ln( 7 4 ) Take ln of both sides. x= 1 2 ln( 7 4 ) Solve for x. 4 e 2x +5=12 4 e 2x =7 Combine like terms. e 2x = 7 4 Divide by the coefficient of the power. 2x=ln( 7 4 ) Take ln of both sides. x= 1 2 ln( 7 4 ) Solve for x.

Try It:

Exercise 7

Solve 3+ e 2t =7 e 2t . 3+ e 2t =7 e 2t .

Solution

t=ln( 1 2 )= 1 2 ln( 2 ) t=ln( 1 2 )= 1 2 ln( 2 )

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution, which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example 8

Problem 1
Solving Exponential Functions in Quadratic Form

Solve e 2x e x =56. e 2x e x =56.

Solution
e 2x e x =56 e 2x e x 56 =0 Get one side of the equation equal to zero. ( e x +7)( e x 8) =0 Factor by the FOIL method. e x +7 =0 or  e x 8=0 If a product is zero, then one factor must be zero. e x =7  or e x =8 Isolate the exponentials. e x =8 Reject the equation in which the power equals a negative number. x =ln8 Solve the equation in which the power equals a positive number. e 2x e x =56 e 2x e x 56 =0 Get one side of the equation equal to zero. ( e x +7)( e x 8) =0 Factor by the FOIL method. e x +7 =0 or  e x 8=0 If a product is zero, then one factor must be zero. e x =7  or e x =8 Isolate the exponentials. e x =8 Reject the equation in which the power equals a negative number. x =ln8 Solve the equation in which the power equals a positive number.
Analysis

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation e x =−7 e x =−7 because a positive number never equals a negative number. The solution ln(−7) ln(−7) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Try It:

Exercise 8

Solve e 2x = e x +2. e 2x = e x +2.

Solution

x=ln2 x=ln2

Q&A:

Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation log b ( x )=y log b ( x )=y is equivalent to the exponential equation b y =x. b y =x. We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation log 2 ( 2 )+ log 2 ( 3x5 )=3. log 2 ( 2 )+ log 2 ( 3x5 )=3. To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for x: x:

log 2 (2)+ log 2 (3x5)=3             log 2 (2(3x5))=3 Apply the product rule of logarithms.                log 2 (6x10)=3 Distribute.                                 2 3 =6x10 Apply the definition of a logarithm.                                   8=6x10 Calculate  2 3 .                                 18=6x Add 10 to both sides.                                  x=3 Divide by 6. log 2 (2)+ log 2 (3x5)=3             log 2 (2(3x5))=3 Apply the product rule of logarithms.                log 2 (6x10)=3 Distribute.                                 2 3 =6x10 Apply the definition of a logarithm.                                   8=6x10 Calculate  2 3 .                                 18=6x Add 10 to both sides.                                  x=3 Divide by 6.

A General Note: Using the Definition of a Logarithm to Solve Logarithmic Equations:

For any algebraic expression S S and real numbers b b and c, c, where b>0, b1, b>0, b1,

log b (S)=cif and only if b c =S log b (S)=cif and only if b c =S
(12)

Example 9

Problem 1

Using Algebra to Solve a Logarithmic Equation

Solve 2lnx+3=7. 2lnx+3=7.

Solution
2lnx+3=7      2lnx=4 Subtract 3.        lnx=2 Divide by 2.            x= e 2 Rewrite in exponential form. 2lnx+3=7      2lnx=4 Subtract 3.        lnx=2 Divide by 2.            x= e 2 Rewrite in exponential form.

Try It:

Exercise 9

Solve 6+lnx=10. 6+lnx=10.

Solution

x= e 4 x= e 4

Example 10

Problem 1

Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve 2ln(6x)=7. 2ln(6x)=7.

Solution
2ln(6x)=7   ln(6x)= 7 2 Divide by 2.         6x= e ( 7 2 ) Use the definition of ln.           x= 1 6 e ( 7 2 ) Divide by 6. 2ln(6x)=7   ln(6x)= 7 2 Divide by 2.         6x= e ( 7 2 ) Use the definition of ln.           x= 1 6 e ( 7 2 ) Divide by 6.

Try It:

Exercise 10

Solve 2ln(x+1)=10. 2ln(x+1)=10.

Solution

x= e 5 1 x= e 5 1

Example 11

Problem 1

Using a Graph to Understand the Solution to a Logarithmic Equation

Solve lnx=3. lnx=3.

Solution
lnx=3 x= e 3 Use the definition of the natural logarithm. lnx=3 x= e 3 Use the definition of the natural logarithm.

Figure 3 represents the graph of the equation. On the graph, the x-coordinate of the point at which the two graphs intersect is close to 20. In other words e 3 20. e 3 20. A calculator gives a better approximation: e 3 20.0855. e 3 20.0855.

Figure 3: The graphs of y=lnx y=lnx and y=3 y=3 cross at the point (e 3 ,3), (e 3 ,3), which is approximately (20.0855, 3).
Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

Try It:

Exercise 11

Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2 x =1000 2 x =1000 to 2 decimal places.

Solution

x9.97 x9.97

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers x>0, x>0, S>0, S>0, T>0 T>0 and any positive real number b, b, where b1, b1,

log b S= log b T if and only if S=T. log b S= log b T if and only if S=T.

For example,

If   log 2 (x1)= log 2 (8),then x1=8. If   log 2 (x1)= log 2 (8),then x1=8.

So, if x1=8, x1=8, then we can solve for x, x, and we get x=9. x=9. To check, we can substitute x=9 x=9 into the original equation: log 2 ( 91 )= log 2 ( 8 )=3. log 2 ( 91 )= log 2 ( 8 )=3. In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation log( 3x2 )log( 2 )=log( x+4 ). log( 3x2 )log( 2 )=log( x+4 ). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for x: x:

log(3x2)log(2)=log(x+4)             log( 3x2 2 )=log(x+4) Apply the quotient rule of logarithms.                      3x2 2 =x+4 Apply the one to one property of a logarithm.                     3x2=2x+8 Multiply both sides of the equation by 2.                              x=10 Subtract 2x and add 2. log(3x2)log(2)=log(x+4)             log( 3x2 2 )=log(x+4) Apply the quotient rule of logarithms.                      3x2 2 =x+4 Apply the one to one property of a logarithm.                     3x2=2x+8 Multiply both sides of the equation by 2.                              x=10 Subtract 2x and add 2.

To check the result, substitute x=10 x=10 into log( 3x2 )log( 2 )=log( x+4 ). log( 3x2 )log( 2 )=log( x+4 ).

log(3(10)2)log(2)=log((10)+4)           log(28)log(2)=log(14)                        log( 28 2 )=log(14) The solution checks. log(3(10)2)log(2)=log((10)+4)           log(28)log(2)=log(14)                        log( 28 2 )=log(14) The solution checks.

A General Note: Using the One-to-One Property of Logarithms to Solve Logarithmic Equations:

For any algebraic expressions S S and T T and any positive real number b, b, where b1, b1,

log b S= log b Tif and only ifS=T log b S= log b Tif and only ifS=T
(20)

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How To:

Given an equation containing logarithms, solve it using the one-to-one property.

  1. Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form log b S= log b T. log b S= log b T.
  2. Use the one-to-one property to set the arguments equal.
  3. Solve the resulting equation, S=T, S=T, for the unknown.

Example 12

Problem 1

Solving an Equation Using the One-to-One Property of Logarithms

Solve ln( x 2 )=ln(2x+3). ln( x 2 )=ln(2x+3).

Solution
          ln( x 2 )=ln(2x+3)                   x 2 =2x+3 Use the one-to-one property of the logarithm.      x 2 2x3=0 Get zero on one side before factoring. (x3)(x+1)=0 Factor using FOIL.               x3=0 or x+1=0 If a product is zero, one of the factors must be zero.                    x=3 or x=1 Solve for x.           ln( x 2 )=ln(2x+3)                   x 2 =2x+3 Use the one-to-one property of the logarithm.      x 2 2x3=0 Get zero on one side before factoring. (x3)(x+1)=0 Factor using FOIL.               x3=0 or x+1=0 If a product is zero, one of the factors must be zero.                    x=3 or x=1 Solve for x.
Analysis

There are two solutions: 3 3 or −1. −1. The solution −1 −1 is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Try It:

Exercise 12

Solve ln( x 2 )=ln1. ln( x 2 )=ln1.

Solution

x=1 x=1 or x=1 x=1

Solving Applied Problems Using Exponential and Logarithmic Equations

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life. Table 1 lists the half-life for several of the more common radioactive substances.

Table 1
Substance Use Half-life
gallium-67 nuclear medicine 80 hours
cobalt-60 manufacturing 5.3 years
technetium-99m nuclear medicine 6 hours
americium-241 construction 432 years
carbon-14 archeological dating 5,715 years
uranium-235 atomic power 703,800,000 years

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

A(t)= A 0 e ln(0.5) T t A(t)= A 0 e ln(0.5) t T A(t)= A 0 ( e ln(0.5) ) t T A(t)= A 0 ( 1 2 ) t T A(t)= A 0 e ln(0.5) T t A(t)= A 0 e ln(0.5) t T A(t)= A 0 ( e ln(0.5) ) t T A(t)= A 0 ( 1 2 ) t T

where

  • A 0 A 0 is the amount initially present
  • T T is the half-life of the substance
  • t t is the time period over which the substance is studied
  • y y is the amount of the substance present after time t t

Example 13

Problem 1

Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a 1000-gram sample of uranium-235 to decay?

Solution
         y=1000e ln(0.5) 703,800,000 t      900=1000 e ln(0.5) 703,800,000 t After 10% decays, 900 grams are left.       0.9= e ln(0.5) 703,800,000 t Divide by 1000. ln(0.9)=ln( e ln(0.5) 703,800,000 t ) Take ln of both sides. ln(0.9)= ln(0.5) 703,800,000 t ln( e M )=M          t=703,800,000× ln(0.9) ln(0.5) years Solve for t.          t106,979,777 years          y=1000e ln(0.5) 703,800,000 t      900=1000 e ln(0.5) 703,800,000 t After 10% decays, 900 grams are left.       0.9= e ln(0.5) 703,800,000 t Divide by 1000. ln(0.9)=ln( e ln(0.5) 703,800,000 t ) Take ln of both sides. ln(0.9)= ln(0.5) 703,800,000 t ln( e M )=M          t=703,800,000× ln(0.9) ln(0.5) years Solve for t.          t106,979,777 years
Analysis

Ten percent of 1000 grams is 100 grams. If 100 grams decay, the amount of uranium-235 remaining is 900 grams.

Try It:

Exercise 13

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

Solution

t=703,800,000× ln(0.8) ln(0.5)  years  226,572,993 years. t=703,800,000× ln(0.8) ln(0.5)  years  226,572,993 years.

Media:

Access these online resources for additional instruction and practice with exponential and logarithmic equations.

Key Equations

Table 2
One-to-one property for exponential functions For any algebraic expressions  S   S  and  T   T  and any positive real number  b,   b,  where
b S = b T   b S = b T   if and only if  S=T.  S=T.
Definition of a logarithm For any algebraic expression S and positive real numbers  b    b   and  c,   c,  where  b1,  b1,
log b (S)=c  log b (S)=c  if and only if   b c =S.   b c =S.
One-to-one property for logarithmic functions For any algebraic expressions S and T and any positive real number  b,   b,  where  b1,  b1,
log b S= log b T  log b S= log b T  if and only if  S=T.  S=T.

Key Concepts

  • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
  • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example 1.
  • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example 2, Example 3, and Example 4.
  • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example 5.
  • We can solve exponential equations with base e, e, by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example 6 and Example 7.
  • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example 8.
  • When given an equation of the form log b (S)=c, log b (S)=c, where S S is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation b c =S, b c =S, and solve for the unknown. See Example 9 and Example 10.
  • We can also use graphing to solve equations with the form log b (S)=c. log b (S)=c. We graph both equations y= log b (S) y= log b (S) and y=c y=c on the same coordinate plane and identify the solution as the x-value of the intersecting point. See Example 11.
  • When given an equation of the form log b S= log b T, log b S= log b T, where S S and T T are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation S=T S=T for the unknown. See Example 12.
  • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example 13.

Section Exercises

Verbal

Exercise 14

How can an exponential equation be solved?

Solution

Determine first if the equation can be rewritten so that each side uses the same base. If so, the exponents can be set equal to each other. If the equation cannot be rewritten so that each side uses the same base, then apply the logarithm to each side and use properties of logarithms to solve.

Exercise 15

When does an extraneous solution occur? How can an extraneous solution be recognized?

Exercise 16

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

Solution

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

Algebraic

For the following exercises, use like bases to solve the exponential equation.

Exercise 17

4 3v2 = 4 v 4 3v2 = 4 v

Exercise 18

64 4 3x =16 64 4 3x =16

Solution

x= 1 3 x= 1 3

Exercise 19

3 2x+1 3 x =243 3 2x+1 3 x =243

Exercise 20

2 3n 1 4 = 2 n+2 2 3n 1 4 = 2 n+2

Solution

n=1 n=1

Exercise 21

625 5 3x+3 =125 625 5 3x+3 =125

Exercise 22

36 3b 36 2b = 216 2b 36 3b 36 2b = 216 2b

Solution

b= 6 5 b= 6 5

Exercise 23

( 1 64 ) 3n 8= 2 6 ( 1 64 ) 3n 8= 2 6

For the following exercises, use logarithms to solve.

Exercise 24

9 x10 =1 9 x10 =1

Solution

x=10 x=10

Exercise 25

2 e 6x =13 2 e 6x =13

Exercise 26

e r+10 10=−42 e r+10 10=−42

Solution

No solution

Exercise 27

2 10 9a =29 2 10 9a =29

Exercise 28

8 10 p+7 7=−24 8 10 p+7 7=−24

Solution

p=log( 17 8 )7 p=log( 17 8 )7

Exercise 29

7 e 3n5 +5=−89 7 e 3n5 +5=−89

Exercise 30

e 3k +6=44 e 3k +6=44

Solution

k= ln( 38 ) 3 k= ln( 38 ) 3

Exercise 31

5 e 9x8 8=−62 5 e 9x8 8=−62

Exercise 32

6 e 9x+8 +2=−74 6 e 9x+8 +2=−74

Solution

x= ln( 38 3 )8 9 x= ln( 38 3 )8 9

Exercise 33

2 x+1 = 5 2x1 2 x+1 = 5 2x1

Exercise 34

e 2x e x 132=0 e 2x e x 132=0

Solution

x=ln12  x=ln12 

Exercise 35

7 e 8x+8 5=−95 7 e 8x+8 5=−95

Exercise 36

10 e 8x+3 +2=8 10 e 8x+3 +2=8

Solution

x= ln( 3 5 )3 8 x= ln( 3 5 )3 8

Exercise 37

4 e 3x+3 7=53 4 e 3x+3 7=53

Exercise 38

8 e 5x2 4=−90 8 e 5x2 4=−90

Solution

no solution

Exercise 39

3 2x+1 = 7 x2 3 2x+1 = 7 x2

Exercise 40

e 2x e x 6=0 e 2x e x 6=0

Solution

x=ln( 3 ) x=ln( 3 )

Exercise 41

3 e 33x +6=−31 3 e 33x +6=−31

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

Exercise 42

log( 1 100 )=−2 log( 1 100 )=−2

Solution

10 2 = 1 100 10 2 = 1 100

Exercise 43

log 324 ( 18 )= 1 2 log 324 ( 18 )= 1 2

For the following exercises, use the definition of a logarithm to solve the equation.

Exercise 44

5 log 7 n=10 5 log 7 n=10

Solution

n=49 n=49

Exercise 45

8 log 9 x=16 8 log 9 x=16

Exercise 46

4+ log 2 ( 9k )=2 4+ log 2 ( 9k )=2

Solution

k= 1 36 k= 1 36

Exercise 47

2log( 8n+4 )+6=10 2log( 8n+4 )+6=10

Exercise 48

104ln( 98x )=6 104ln( 98x )=6

Solution

x= 9e 8 x= 9e 8

For the following exercises, use the one-to-one property of logarithms to solve.

Exercise 49

ln( 103x )=ln( 4x ) ln( 103x )=ln( 4x )

Exercise 50

log 13 ( 5n2 )= log 13 ( 85n ) log 13 ( 5n2 )= log 13 ( 85n )

Solution

n=1 n=1

Exercise 51

log( x+3 )log( x )=log( 74 ) log( x+3 )log( x )=log( 74 )

Exercise 52

ln( 3x )=ln( x 2 6x ) ln( 3x )=ln( x 2 6x )

Solution

No solution

Exercise 53

log 4 ( 6m )= log 4 3m log 4 ( 6m )= log 4 3m

Exercise 54

ln( x2 )ln( x )=ln( 54 ) ln( x2 )ln( x )=ln( 54 )

Solution

No solution

Exercise 55

log 9 ( 2 n 2 14n )= log 9 ( 45+ n 2 ) log 9 ( 2 n 2 14n )= log 9 ( 45+ n 2 )

Exercise 56

ln( x 2 10 )+ln( 9 )=ln( 10 ) ln( x 2 10 )+ln( 9 )=ln( 10 )

Solution

x=± 10 3 x=± 10 3

For the following exercises, solve each equation for x. x.

Exercise 57

log(x+12)=log(x)+log(12) log(x+12)=log(x)+log(12)

Exercise 58

ln(x)+ln(x3)=ln(7x) ln(x)+ln(x3)=ln(7x)

Solution

x=10 x=10

Exercise 59

log 2 (7x+6)=3 log 2 (7x+6)=3

Exercise 60

ln( 7 )+ln( 24 x 2 )=ln( 14 ) ln( 7 )+ln( 24 x 2 )=ln( 14 )

Solution

x=0 x=0

Exercise 61

log 8 ( x+6 ) log 8 ( x )= log 8 ( 58 ) log 8 ( x+6 ) log 8 ( x )= log 8 ( 58 )

Exercise 62

ln( 3 )ln( 33x )=ln( 4 ) ln( 3 )ln( 33x )=ln( 4 )

Solution

x= 3 4 x= 3 4

Exercise 63

log 3 ( 3x ) log 3 ( 6 )= log 3 ( 77 ) log 3 ( 3x ) log 3 ( 6 )= log 3 ( 77 )

Graphical

For the following exercises, solve the equation for x, x, if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

Exercise 64

log 9 ( x )5=−4 log 9 ( x )5=−4

Solution

x=9 x=9

Graph of log_9(x)-5=y and y=-4.

Exercise 65

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

Exercise 66

ln( 3x )=2 ln( 3x )=2

Solution

x= e 2 3 2.5 x= e 2 3 2.5

Graph of ln(3x)=y and y=2.

Exercise 67

ln( x5 )=1 ln( x5 )=1

Exercise 68

log( 4 )+log( 5x )=2 log( 4 )+log( 5x )=2

Solution

x=5 x=5

Graph of log(4)+log(-5x)=y and y=2.

Exercise 69

7+ log 3 ( 4x )=−6 7+ log 3 ( 4x )=−6

Exercise 70

ln( 4x10 )6=5 ln( 4x10 )6=5

Solution

x= e+10 4 3.2 x= e+10 4 3.2

Graph of ln(4x-10)-6=y and y=-5.

Exercise 71

log( 42x )=log( 4x ) log( 42x )=log( 4x )

Exercise 72

log 11 ( 2 x 2 7x )= log 11 ( x2 ) log 11 ( 2 x 2 7x )= log 11 ( x2 )

Solution

No solution

Graph of log_11(-2x^2-7x)=y and y=log_11(x-2).

Exercise 73

ln( 2x+9 )=ln( 5x ) ln( 2x+9 )=ln( 5x )

Exercise 74

log 9 ( 3x )= log 9 ( 4x8 ) log 9 ( 3x )= log 9 ( 4x8 )

Solution

x= 11 5 2.2 x= 11 5 2.2

Graph of log_9(3-x)=y and y=log_9(4x-8).

Exercise 75

log( x 2 +13 )=log( 7x+3 ) log( x 2 +13 )=log( 7x+3 )

Exercise 76

3 log 2 ( 10 ) log( x9 )=log( 44 ) 3 log 2 ( 10 ) log( x9 )=log( 44 )

Solution

x= 101 11 9.2 x= 101 11 9.2

Graph of 3/log_2(10)-log(x-9)=y and y=log(44).

Exercise 77

ln( x )ln( x+3 )=ln( 6 ) ln( x )ln( x+3 )=ln( 6 )

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

Exercise 78

An account with an initial deposit of $6,500 $6,500 earns 7.25% 7.25% annual interest, compounded continuously. How much will the account be worth after 20 years?

Solution

about $27,710.24 $27,710.24

Graph of f(x)=6500e^(0.0725x) with the labeled point at (20, 27710.24).

Exercise 79

The formula for measuring sound intensity in decibels D D is defined by the equation D=10log( I I 0 ), D=10log( I I 0 ), where I I is the intensity of the sound in watts per square meter and I 0 = 10 12 I 0 = 10 12 is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of 8.3 10 2 8.3 10 2 watts per square meter?

Exercise 80

The population of a small town is modeled by the equation P=1650 e 0.5t P=1650 e 0.5t where t t is measured in years. In approximately how many years will the town’s population reach 20,000? 20,000?

Solution

about 5 years

Graph of P(t)=1650e^(0.5x) with the labeled point at (5, 20000).

Technology

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate x x to 3 decimal places.

Exercise 81

1000 ( 1.03 ) t =5000 1000 ( 1.03 ) t =5000 using the common log.

Exercise 82

e 5x =17 e 5x =17 using the natural log

Solution

ln(17) 5 0.567 ln(17) 5 0.567

Exercise 83

3 ( 1.04 ) 3t =8 3 ( 1.04 ) 3t =8 using the common log

Exercise 84

3 4x5 =38 3 4x5 =38 using the common log

Solution

x= log( 38 )+5log( 3 )    4log( 3 ) 2.078 x= log( 38 )+5log( 3 )    4log( 3 ) 2.078

Exercise 85

50 e 0.12t =10 50 e 0.12t =10 using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

Exercise 86

7 e 3x5 +7.9=47 7 e 3x5 +7.9=47

Solution

x2.2401 x2.2401

Exercise 87

ln( 3 )+ln( 4.4x+6.8 )=2 ln( 3 )+ln( 4.4x+6.8 )=2

Exercise 88

log( 0.7x9 )=1+5log( 5 ) log( 0.7x9 )=1+5log( 5 )

Solution

x44655.7143 x44655.7143

Exercise 89

Atmospheric pressure P P in pounds per square inch is represented by the formula P=14.7 e 0.21x , P=14.7 e 0.21x , where x x is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of 8.369 8.369 pounds per square inch? (Hint: there are 5280 feet in a mile)

Exercise 90

The magnitude M of an earthquake is represented by the equation M= 2 3 log( E E 0 ) M= 2 3 log( E E 0 ) where E E is the amount of energy released by the earthquake in joules and E 0 = 10 4.4 E 0 = 10 4.4 is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing 1.4 10 13 1.4 10 13 joules of energy?

Solution

about 5.83 5.83

Extensions

Exercise 91

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that b log b x =x. b log b x =x.

Exercise 92

Recall the formula for continually compounding interest, y=A e kt . y=A e kt . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

Solution

t=ln( ( y A ) 1 k ) t=ln( ( y A ) 1 k )

Exercise 93

Recall the compound interest formula A=a ( 1+ r k ) kt . A=a ( 1+ r k ) kt . Use the definition of a logarithm along with properties of logarithms to solve the formula for time t. t.

Exercise 94

Newton’s Law of Cooling states that the temperature T T of an object at any time t can be described by the equation T= T s +( T 0 T s ) e kt , T= T s +( T 0 T s ) e kt , where T s T s is the temperature of the surrounding environment, T 0 T 0 is the initial temperature of the object, and k k is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time t t such that t t is equal to a single logarithm.

Solution

t=ln( ( T T s T 0 T s ) 1 k ) t=ln( ( T T s T 0 T s ) 1 k )

Glossary

extraneous solution:
a solution introduced while solving an equation that does not satisfy the conditions of the original equation

Content actions

Download module as:

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks