12.6: Solve Exponential and Logarithmic Equations
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Learning Objectives
By the end of this section, you will be able to:
- Solve logarithmic equations using the properties of logarithms
- Solve exponential equations using logarithms
- Use exponential models in applications
Before you get started, take this readiness quiz.
- Solve: x2=16.
If you missed this problem, review Example 6.46. - Solve: x2−5x+6=0.
If you missed this problem, review Example 6.45. - Solve: x(x+6)=2x+5.
If you missed this problem, review Example 6.47.
Solve Logarithmic Equations Using the Properties of Logarithms
In the section on logarithmic functions, we solved some equations by rewriting the equation in exponential form. Now that we have the properties of logarithms, we have additional methods we can use to solve logarithmic equations.
If our equation has two logarithms we can use a property that says that if logaM=logaN then it is true that M=N. This is the One-to-One Property of Logarithmic Equations.
Definition 12.6.1
One-to-One Property of Logarithmic Equations
For M>0,N>0,a>0, and a≠1 is any real number:
If logaM=logaN, then M=N.
To use this property, we must be certain that both sides of the equation are written with the same base.
Remember that logarithms are defined only for positive real numbers. Check your results in the original equation. You may have obtained a result that gives a logarithm of zero or a negative number.
Example 12.6.1
Solve: 2log5x=log581.
Solution:
2log5x=log581
Use the Power Property.
log5x2=log581
Use the One-to-One Property, if logaM=logaN, then M=N.
x2=81
Solve using the Square Root Property.
x=±9
We eliminate x=−9 as we cannot take the logarithm of a negative number.
x=9,x=−9
Check. x=9
2log5x=log5812log59?=log581log592?=log581log581=log581
Exercise 12.6.1
Solve: 2log3x=log336
- Answer
-
x=6
Exercise 12.6.2
Solve: 3logx=log64
- Answer
-
x=4
Another strategy to use to solve logarithmic equations is to condense sums or differences into a single logarithm.
Example 12.6.2
Solve: log3x+log3(x−8)=2.
Solution:
log3x+log3(x−8)=2
Use the Product Property, logaM+logaN=logaM⋅N.
log3x(x−8)=2
Rewrite in exponential form.
32=x(x−8)
Simplify.
9=x2−8x
Subtract 9 from each side.
0=x2−8x−9
Factor.
0=(x−9)(x+1)
Use the Zero-Product-Property
x−9=0,x+1=0
Solve each equation.
x=9,x=−1
Check. x=−1
log3x+log3(x−8)=2log3(−1)+log3(−1−8)?=2
We cannot take the log of a negative number.
Check. x=9
log3x+log3(x−8)=2log39+log3(9−8)?=22+0?=22=2
Exercise 12.6.3
Solve: log2x+log2(x−2)=3
- Answer
-
x=4
Exercise 12.6.4
Solve: log2x+log2(x−6)=4
- Answer
-
x=8
When there are logarithms on both sides, we condense each side into a single logarithm. Remember to use the Power Property as needed.
Example 12.6.3
Solve: log4(x+6)−log4(2x+5)=−log4x.
Solution:
log4(x+6)−log4(2x+5)=−log4x
Use the Quotient Property on the left side and the PowerProperty on the right.
log4(x+62x+5)=log4x−1
Rewrite x−1=1x.
log4(x+62x+5)=log41x
Use the One-to-One Property, if logaM=logaN, then M=N.
x+62x+5=1x
Solve the rational equation.
x(x+6)=2x+5
Distribute.
x2+6x=2x+5
Write in standard form.
x2+4x−5=0
Factor.
(x+5)(x−1)=0
Use the Zero-Product-Property.
x+5=0,x−1=0
Solve each equation.
x=−5,x=1
Check.
We leave the check for you.
Exercise 12.6.5
Solve: log(x+2)−log(4x+3)=−logx.
- Answer
-
x=3
Exercise 12.6.6
Solve: log(x−2)−log(4x+16)=log1x.
- Answer
-
x=8
Example 12.6.4 Solve Exponential Equations Using Logarithms
Solve 5x=11. Find the exact answer and then approximate it to three decimal places.
Solution:
5x=11
Since the exponential is isolated, take the logarithm of both sides.
log5x=log11
Use the Power Property to get the x as a factor, not an exponent.
xlog5=log11
Solve for x. Find the exact answer.
x=log11log5
Approximate the answer.
x≈1.490
Since 51=5 and 52=25, does it makes sense that 51.490≈11?
Exercise 12.6.7
Solve 7x=43. Find the exact answer and then approximate it to three decimal places.
- Answer
-
x=log43log7≈1.933
Exercise 12.6.8
Solve 8x=98. Find the exact answer and then approximate it to three decimal places.
- Answer
-
x=log98log8≈2.205
When we take the logarithm of both sides we will get the same result whether we use the common or the natural logarithm (try using the natural log in the last example. Did you get the same result?) When the exponential has base e, we use the natural logarithm.
Example 12.6.5
Solve 3ex+2=24. Find the exact answer and then approximate it to three decimal places.
Solution:
3ex+2=24
Isolate the exponential by dividing both sides by 3.
ex+2=8
Take the natural logarithm of both sides.
lnex+2=ln8
Use the Power Property to get the x as a factor, not an exponent.
(x+2)lne=ln8
Use the property lne=1 to simplify.
x+2=ln8
Solve the equation. Find the exact answer.
x=ln8−2
Approximate the answer.
x≈0.079
Exercise 12.6.9
Solve 2ex−2=18. Find the exact answer and then approximate it to three decimal places.
- Answer
-
x=ln9+2≈4.197
Exercise 12.6.10
Solve 5e2x=25. Find the exact answer and then approximate it to three decimal places.
- Answer
-
x=ln52≈0.805
Use Exponential Models in Applications
In previous sections we were able to solve some applications that were modeled with exponential equations. Now that we have so many more options to solve these equations, we are able to solve more applications.
We will again use the Compound Interest Formulas and so we list them here for reference.
Definition 12.6.2
Compound Interest
For a principal, P, invested at an interest rate, r, for t years, the new balance, A is:
A=P(1+rn)nt when compounded n times a year. A=Pert when compounded continuously.
Example 12.6.6
Jermael’s parents put $10,000 in investments for his college expenses on his first birthday. They hope the investments will be worth $50,000 when he turns 18. If the interest compounds continuously, approximately what rate of growth will they need to achieve their goal?
Solution:
Identify the variables in the formula.
A=$50,000P=$10,000r=?t=17 years A=Pert
Substitute the values into the formula.
50,000=10,000er⋅17
Solve for r. Divide each side by 10,000.
5=e17r
Take the natural log of each side.
ln5=lne17r
Use the Power Property.
ln5=17rlne
Simplify.
ln5=17r
Divide each side by 17.
ln517=r
Approximate the answer.
r≈0.095
Convert to a percentage.
r≈9.5%
They need the rate of growth to be approximately 9.5%.
Exercise 12.6.11
Hector invests $10,000 at age 21. He hopes the investments will be worth $150,000 when he turns 50. If the interest compounds continuously, approximately what rate of growth will he need to achieve his goal?
- Answer
-
r≈9.3%
Exercise 12.6.12
Rachel invests $15,000 at age 25. She hopes the investments will be worth $90,000 when she turns 40. If the interest compounds continuously, approximately what rate of growth will she need to achieve her goal?
- Answer
-
r≈11.9%
We have seen that growth and decay are modeled by exponential functions. For growth and decay we use the formula A=A0ekt. Exponential growth has a positive rate of growth or growth constant, k, and exponential decay has a negative rate of growth or decay constant, k.
Definition 12.6.3
Exponential Growth and Decay
For an original amount, A0, that grows or decays at a rate, k, for a certain time, t, the final amount, A, is:
A=A0ekt
We can now solve applications that give us enough information to determine the rate of growth. We can then use that rate of growth to predict other situations.
Example 12.6.7
Researchers recorded that a certain bacteria population grew from 100 to 300 in 3 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?
Solution:
This problem requires two main steps. First we must find the unknown rate, k. Then we use that value of k to help us find the unknown number of bacteria.
Identify the variables in the formula.
A=300A0=100k=?t=3 hours A=A0ekt
Substitute the values in the formula.
300=100ek⋅3
Solve for k. Divide each side by 100.
3=e3k
Take the natural log of each side.
ln3=lne3k
Use the Power Property.
ln3=3klne
Simplify.
ln3=3k
Divide each side by 3.
ln33=k
Approximate the answer.
k≈0.366
We use this rate of growth to predict the number of bacteria there will be in 24 hours.
A=?A0=100k=ln33t=24 hours A=A0ekt
Substitute in the values.
A=100eln33⋅24
Evaluate.
A≈656,100
At this rate of growth, they can expect 656,100 bacteria.
Exercise 12.6.13
Researchers recorded that a certain bacteria population grew from 100 to 500 in 6 hours. At this rate of growth, how many bacteria will there be 24 hours from the start of the experiment?
- Answer
-
There will be 62,500 bacteria.
Exercise 12.6.14
Researchers recorded that a certain bacteria population declined from 700,000 to 400,000 in 5 hours after the administration of medication. At this rate of decay, how many bacteria will there be 24 hours from the start of the experiment?
- Answer
-
There will be 5,870,061 bacteria.
Radioactive substances decay or decompose according to the exponential decay formula. The amount of time it takes for the substance to decay to half of its original amount is called the half-life of the substance.
Similar to the previous example, we can use the given information to determine the constant of decay, and then use that constant to answer other questions.
Example 12.6.8
The half-life of radium-226 is 1,590 years. How much of a 100 mg sample will be left in 500 years?
Solution:
This problem requires two main steps. First we must find the decay constant k. If we start with 100-mg, at the half-life there will be 50-mg remaining. We will use this information to find k. Then we use that value of k to help us find the amount of sample that will be left in 500 years.
Identify the variables in the formula.
A=50A0=100k=?t=1590 years A=A0ekt
Substitute the values in the formula.
50=100ek⋅1590
Solve for k. Divide each side by 100.
0.5=e1590k
Take the natural log of each side.
ln0.5=lne1590k
Use the Power Property.
ln0.5=1590klne
Simplify.
ln0.5=1590k
Divide each side by 1590.
ln0.51590=k exact answer
We use this rate of growth to predict the amount that will be left in 500 years.
A=?A0=100k=ln0.51590t=500yearsA=A0ekt
Substitute in the values.
A=100e1n0.51500⋅500
Evaluate.
A≈80.4mg
In 500 years there would be approximately 80.4 mg remaining.
Exercise 12.6.15
The half-life of magnesium-27 is 9.45 minutes. How much of a 10-mg sample will be left in 6 minutes?
- Answer
-
There will be 6.43 mg left.
Exercise 12.6.16
The half-life of radioactive iodine is 60 days. How much of a 50-mg sample will be left in 40 days?
- Answer
-
There will be 31.5 mg left.
Access these online resources for additional instruction and practice with solving exponential and logarithmic equations.
Key Concepts
- One-to-One Property of Logarithmic Equations: For M>0,N>0,a>0, and a≠1 is any real number:
If logaM=logaN, then M=N
- Compound Interest:
For a principal, P, invested at an interest rate, r, for t years, the new balance, A, is:A=P(1+rn)nt when compounded n times a year. A=Pert when compounded continuously.
- Exponential Growth and Decay: For an original amount, A0 that grows or decays at a rate, r, for a certain time t, the final amount, A, is A=A0ert.