Exponential and Logarithmic Equations

Learning Objectives

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.

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)

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,b≠1,𝑏>0,𝑏≠1, bS=bT𝑏𝑆=𝑏𝑇 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 34x−7=32×3.34𝑥−7=32𝑥3. 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 x𝑥:

34x−734x−734x−74x−72xx=32×3=32×31=32x−1=2x−1=6=3Rewrite 3 as 31.Use the division property of exponents.Apply the one-to-one property of exponents.Subtract 2xand add 7 to both sides.Divide by 2.34𝑥−7=32𝑥334𝑥−7=32𝑥31Rewrite 3 as 31.34𝑥−7=32𝑥−1Use the division property of exponents.4𝑥−7=2𝑥−1Apply the one-to-one property of exponents.2𝑥=6Subtract 2𝑥and add 7 to both sides.𝑥=3Divide by 2.

USING THE ONE-TO-ONE PROPERTY OF EXPONENTIAL FUNCTIONS TO SOLVE EXPONENTIAL EQUATIONS

For any algebraic expressions Sand T,𝑆and 𝑇, and any positive real number b≠1,𝑏≠1,

bS=bTif and only ifS=T𝑏𝑆=𝑏𝑇if and only if𝑆=𝑇

HOW TO

Given an exponential equation with the form bS=bT,𝑏𝑆=𝑏𝑇, where S𝑆 and T𝑇 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 bS=bT.𝑏𝑆=𝑏𝑇.
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

Solving an Exponential Equation with a Common Base

Solve 2x−1=22x−4.2𝑥−1=22𝑥−4.

Solution

 2x−1=22x−4x−1=2x−4x=3The common base is 2.By the one-to-one property the exponents must be equal.Solve for x. 2𝑥−1=22𝑥−4The common base is 2.𝑥−1=2𝑥−4By the one-to-one property the exponents must be equal.𝑥=3Solve for 𝑥.

TRY IT #1

Solve 52x=53x+2.52𝑥=53𝑥+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=4x−5.256=4𝑥−5. 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=4x−528=(22)x−528=22x−108=2x−1018=2xx=9Rewrite each side as a power with base 2.Use the one-to-one property of exponents.Apply the one-to-one property of exponents.Add 10 to both sides.Divide by 2.256=4𝑥−528=(22)𝑥−5Rewrite each side as a power with base 2.28=22𝑥−10Use the one-to-one property of exponents.8=2𝑥−10Apply the one-to-one property of exponents.18=2𝑥Add 10 to both sides.𝑥=9Divide 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 bS=bT.𝑏𝑆=𝑏𝑇.
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

Solving Equations by Rewriting Them to Have a Common Base

Solve 8x+2=16x+1.8𝑥+2=16𝑥+1.

Solution

8x+2=16x+1(23)x+2=(24)x+123x+6=24x+43x+6=4x+4x=2Write8and16as powers of2.To take a power of a power, multiply exponents.Use the one-to-one property to set the exponents equal.Solve for x.8𝑥+2=16𝑥+1(23)𝑥+2=(24)𝑥+1Write8and16as powers of2.23𝑥+6=24𝑥+4To take a power of a power, multiply exponents.3𝑥+6=4𝑥+4Use the one-to-one property to set the exponents equal.𝑥=2Solve for 𝑥.

TRY IT #2

Solve 52x=253x+2.52𝑥=253𝑥+2.

EXAMPLE 3

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

Solve 25x=2–√.25𝑥=2.

Solution

25x=2125x=12x=110Write the square root of  2 as a power of2.Use the one-to-one property.Solve forx.25𝑥=212Write the square root of  2 as a power of2.5𝑥=12Use the one-to-one property.𝑥=110Solve for𝑥.

TRY IT #3

Solve 5x=5–√.5𝑥=5.

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

Solving an Equation with Positive and Negative Powers

Solve 3x+1=−2.3𝑥+1=−2.

Solution

This equation has no solution. There is no real value of x𝑥 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

TRY IT #4

Solve 2x=−100.2𝑥=−100.

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.
   1. If one of the terms in the equation has base 10, use the common logarithm.
   2. 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

Solving an Equation Containing Powers of Different Bases

Solve 5x+2=4x.5𝑥+2=4𝑥.

Solution

5x+2=4xln5x+2=ln4x(x+2)ln5=xln4xln5+2ln5=xln4xln5−xln4=−2ln5x(ln5−ln4)=−2ln5xln(54)=ln(125)x=ln(125)ln(54)There is no easy way to get the powers to have the same base.Take ln of both sides.Use laws of logs.Use the distributive law.Get terms containingxon one side, terms withoutxon the other.On the left hand side, factor out anx.Use the laws of logs.Divide by the coefficient ofx.5𝑥+2=4𝑥There is no easy way to get the powers to have the same base.ln5𝑥+2=ln4𝑥Take ln of both sides.(𝑥+2)ln5=𝑥ln4Use laws of logs.𝑥ln5+2ln5=𝑥ln4Use the distributive law.𝑥ln5−𝑥ln4=−2ln5Get terms containing𝑥on one side, terms without𝑥on the other.𝑥(ln5−ln4)=−2ln5On the left hand side, factor out an𝑥.𝑥ln(54)=ln(125)Use the laws of logs.𝑥=ln(125)ln(54)Divide by the coefficient of𝑥.

TRY IT #5

Solve 2x=3x+1.2𝑥=3𝑥+1.

Q&A

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

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 e𝑒 on either side, we can use the natural logarithm to solve it.

HOW TO

Given an equation of the form y=Aekt,𝑦=𝐴𝑒𝑘𝑡, 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

Solve an Equation of the Form y = Ae^kt

Solve 100=20e2t.100=20𝑒2𝑡.

Solution

1005ln5t=20e2t=e2t=2t=ln52Divide by the coefficient of the power.Take ln of both sides. Use the fact that ln(x)and exare inverse functions.Divide by the coefficient of t.100=20𝑒2𝑡5=𝑒2𝑡Divide by the coefficient of the power.ln5=2𝑡Take ln of both sides. Use the fact that ln(𝑥)and 𝑒𝑥are inverse functions.𝑡=ln52Divide by the coefficient of 𝑡.

Analysis

Using laws of logs, we can also write this answer in the form t=ln5–√.𝑡=ln5. If we want a decimal approximation of the answer, we use a calculator.

TRY IT #6

Solve 3e0.5t=11.3𝑒0.5𝑡=11.

Q&A

Does every equation of the form y=Aekt𝑦=𝐴𝑒𝑘𝑡 have a solution?

No. There is a solution when k≠0,𝑘≠0, and when y𝑦 and A𝐴 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=−3et.2=−3𝑒𝑡.

EXAMPLE 7

Solving an Equation That Can Be Simplified to the Form y = Ae^kt

Solve 4e2x+5=12.4𝑒2𝑥+5=12.

Solution

4e2x+5=124e2x=7e2x=742x=ln(74)x=12ln(74)Combine like terms.Divide by the coefficient of the power.Take ln of both sides.Solve for x.4𝑒2𝑥+5=124𝑒2𝑥=7Combine like terms.𝑒2𝑥=74Divide by the coefficient of the power.2𝑥=ln(74)Take ln of both sides.𝑥=12ln(74)Solve for 𝑥.

TRY IT #7

Solve 3+e2t=7e2t.3+𝑒2𝑡=7𝑒2𝑡.

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

Solving Exponential Functions in Quadratic Form

Solve e2x−ex=56.𝑒2𝑥−𝑒𝑥=56.

Solution

e2x−exe2x−ex−56(ex+7)(ex−8)ex+7exexx=56=0=0=0orex−8=0=−7or ex=8=8=ln8Get one side of the equation equal to zero.Factor by the FOIL method.If a product is zero, then one factor must be zero.Isolate the exponentials.Reject the equation in which the power equals a negative number.Solve the equation in which the power equals a positive number.𝑒2𝑥−𝑒𝑥=56𝑒2𝑥−𝑒𝑥−56=0Get one side of the equation equal to zero.(𝑒𝑥+7)(𝑒𝑥−8)=0Factor by the FOIL method.𝑒𝑥+7=0or𝑒𝑥−8=0If a product is zero, then one factor must be zero.𝑒𝑥=−7or e𝑥=8Isolate the exponentials.𝑒𝑥=8Reject the equation in which the power equals a negative number.𝑥=ln8Solve 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 ex=−7𝑒𝑥=−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 #8

Solve e2x=ex+2.𝑒2𝑥=𝑒𝑥+2.

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 logb(x)=ylog𝑏(𝑥)=𝑦 is equivalent to the exponential equation by=x.𝑏𝑦=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 log2(2)+log2(3x−5)=3.log2(2)+log2(3𝑥−5)=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:

log2(2)+log2(3x−5)=3log2(2(3x−5))=3log2(6x−10)=323=6x−108=6x−1018=6xx=3Apply the product rule of logarithms.Distribute.Apply the definition of a logarithm.Calculate23.Add 10 to both sides.Divide by 6.log2(2)+log2(3𝑥−5)=3log2(2(3𝑥−5))=3Apply the product rule of logarithms.log2(6𝑥−10)=3Distribute.23=6𝑥−10Apply the definition of a logarithm.8=6𝑥−10Calculate23.18=6𝑥Add 10 to both sides.𝑥=3Divide by 6.

USING THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expression S𝑆 and real numbers b𝑏 and c,𝑐, where b>0,b≠1,𝑏>0,𝑏≠1,

logb(S)=cif and only ifbc=Slog𝑏(𝑆)=𝑐if and only if𝑏𝑐=𝑆

EXAMPLE 9

Using Algebra to Solve a Logarithmic Equation

Solve 2lnx+3=7.2ln𝑥+3=7.

Solution

2lnx+3=72lnx=4lnx=2x=e2Subtract 3.Divide by 2.Rewrite in exponential form.2ln𝑥+3=72ln𝑥=4Subtract 3.ln𝑥=2Divide by 2.𝑥=𝑒2Rewrite in exponential form.

TRY IT #9

Solve 6+lnx=10.6+ln𝑥=10.

EXAMPLE 10

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

Solve 2ln(6x)=7.2ln(6𝑥)=7.

Solution

2ln(6x)=7ln(6x)=726x=e(72)x=16e(72)Divide by 2.Use the definition of ln.Divide by 6.2ln(6𝑥)=7ln(6𝑥)=72Divide by 2.6𝑥=𝑒(72)Use the definition of ln.𝑥=16𝑒(72)Divide by 6.

TRY IT #10

Solve 2ln(x+1)=10.2ln(𝑥+1)=10.

EXAMPLE 11

Using a Graph to Understand the Solution to a Logarithmic Equation

Solve lnx=3.ln𝑥=3.

Solution

lnx=3x=e3Use the definition of the natural logarithm.ln𝑥=3𝑥=𝑒3Use 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 e3≈20.𝑒3≈20. A calculator gives a better approximation: e3≈20.0855.𝑒3≈20.0855.

Figure 3 The graphs of y=lnx𝑦=ln𝑥 and y=3𝑦=3 cross at the point (e3,3),(e3,3), which is approximately (20.0855, 3).

TRY IT #11

Use a graphing calculator to estimate the approximate solution to the logarithmic equation 2x=10002𝑥=1000 to 2 decimal places.

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>0𝑇>0 and any positive real number b,𝑏, where b≠1,𝑏≠1,

logbS=logbTif and only if S=T.log𝑏𝑆=log𝑏𝑇if and only if 𝑆=𝑇.

For example,

If  log2(x−1)=log2(8),then x−1=8.If  log2(𝑥−1)=log2(8),then 𝑥−1=8.

So, if x−1=8,𝑥−1=8, then we can solve for x,𝑥, and we get x=9.𝑥=9. To check, we can substitute x=9𝑥=9 into the original equation: log2(9−1)=log2(8)=3.log2(9−1)=log2(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(3x−2)−log(2)=log(x+4).log(3𝑥−2)−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(3x−2)−log(2)=log(x+4)log(3x−22)=log(x+4)3x−22=x+43x−2=2x+8x=10Apply the quotient rule of logarithms.Apply the one to one property of a logarithm.Multiply both sides of the equation by 2.Subtract 2xand add 2.log(3𝑥−2)−log(2)=log(𝑥+4)log(3𝑥−22)=log(𝑥+4)Apply the quotient rule of logarithms.3𝑥−22=𝑥+4Apply the one to one property of a logarithm.3𝑥−2=2𝑥+8Multiply both sides of the equation by 2.𝑥=10Subtract 2𝑥and add 2.

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

log(3(10)−2)−log(2)=log((10)+4)log(28)−log(2)=log(14)log(282)=log(14)The solution checks.log(3(10)−2)−log(2)=log((10)+4)log(28)−log(2)=log(14)log(282)=log(14)The solution checks.

USING THE ONE-TO-ONE PROPERTY OF LOGARITHMS TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expressions S𝑆 and T𝑇 and any positive real number b,𝑏, where b≠1,𝑏≠1,

logbS=logbTif and only ifS=Tlog𝑏𝑆=log𝑏𝑇if and only if𝑆=𝑇

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 logbS=logbT.log𝑏𝑆=log𝑏𝑇.
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

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

Solve ln(x2)=ln(2x+3).ln(𝑥2)=ln(2𝑥+3).

Solution

ln(x2)=ln(2x+3)x2=2x+3×2−2x−3=0(x−3)(x+1)=0x−3=0or x+1=0x=3orx=−1Use the one-to-one property of the logarithm.Get zero on one side before factoring.Factor using FOIL.If a product is zero, one of the factors must be zero.Solve for x.ln(𝑥2)=ln(2𝑥+3)𝑥2=2𝑥+3Use the one-to-one property of the logarithm.𝑥2−2𝑥−3=0Get zero on one side before factoring.(𝑥−3)(𝑥+1)=0Factor using FOIL.𝑥−3=0or 𝑥+1=0If a product is zero, one of the factors must be zero.𝑥=3or𝑥=−1Solve for 𝑥.

Analysis

There are two solutions: 33 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 #12

Solve ln(x2)=ln1.ln(𝑥2)=ln1.

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.

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)=A0eln(0.5)TtA(t)=A0eln(0.5)tTA(t)=A0(eln(0.5))tTA(t)=A0(12)tT𝐴(𝑡)=𝐴0𝑒ln(0.5)𝑇𝑡𝐴(𝑡)=𝐴0𝑒ln(0.5)𝑡𝑇𝐴(𝑡)=𝐴0(𝑒ln(0.5))𝑡𝑇𝐴(𝑡)=𝐴0(12)𝑡𝑇

where

• A0𝐴0 is the amount initially present
• T𝑇 is the half-life of the substance
• t𝑡 is the time period over which the substance is studied
• A(t)𝐴(𝑡) is the amount of the substance present after time t𝑡

EXAMPLE 13

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=1000eln(0.5)703,800,000t900=1000eln(0.5)703,800,000t0.9=eln(0.5)703,800,000tln(0.9)=ln(eln(0.5)703,800,000t)ln(0.9)=ln(0.5)703,800,000tt=703,800,000×ln(0.9)ln(0.5)yearst≈106,979,777 yearsAfter 10% decays, 900 grams are left.Divide by 1000.Take ln of both sides.ln(eM)=MSolve fort.𝑦=1000𝑒ln(0.5)703,800,000𝑡900=1000𝑒ln(0.5)703,800,000𝑡After 10% decays, 900 grams are left.0.9=𝑒ln(0.5)703,800,000𝑡Divide by 1000.ln(0.9)=ln(𝑒ln(0.5)703,800,000𝑡)Take ln of both sides.ln(0.9)=ln(0.5)703,800,000𝑡ln(𝑒𝑀)=𝑀𝑡=703,800,000×ln(0.9)ln(0.5)yearsSolve for𝑡.𝑡≈106,979,777 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 #13

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