College AlgebraFifth EditionJames Stewart Lothar Redlin Saleem Watson

Exponential and Logarithmic Functions5

Exponential and Logarithmic Equations5.4

Exponential and Logarithmic Equations

In this section, we solve equations

that involve exponential or logarithmic

functions.

• The techniques we develop here will be used in the next section for solving applied problems.

Exponential Equations

Exponential Equations

An exponential equation is one in which

the variable occurs in the exponent.

• For example, 2x = 7

Solving Exponential Equations

The variable x presents a difficulty

because it is in the exponent.

• To deal with this difficulty, we take the logarithm of each side and then use the Laws of Logarithms to “bring down x” from the exponent.

Solving Exponential Equations

• Law 3 of the Laws of Logarithms says that: logaAC = C logaA

(Law 3)

2 7

ln2 ln7

ln2 ln7

ln72.807

ln2

x

x

x

x

Solving Exponential Equations

The method we used to solve 2x = 7

is typical of how we solve exponential

equations in general.

Guidelines for Solving Exponential Equations

1. Isolate the exponential expression on

one side of the equation.

2. Take the logarithm of each side,

and then use the Laws of Logarithms

to “bring down the exponent.”

3. Solve for the variable.

E.g. 1—Solving an Exponential Equation

Find the solution of

3x + 2 = 7

correct to six decimal places.

• We take the common logarithm of each side and use Law 3.

E.g. 1—Solving an Exponential Equation

2

2

(Law 3)

3 7

log 3 log7

2 log3 log7

log72

log3

log72 0.228756

log3

x

x

x

x

x

Check Your Answer

Substituting x = –0.228756 into

the original equation and using

a calculator, we get:

3(–0.228756)+2 ≈ 7

E.g. 2—Solving an Exponential Equation

Solve the equation 8e2x = 20.

• We first divide by 8 to isolate the exponential term on one side.

2

2 208

2

8 20

ln ln2.5

2 ln2.5

ln2.50.458

2

x

x

x

e

e

e

x

x

Check Your Answer

Substituting x = 0.458 into the original

equation and using a calculator,

we get:

8e2(0.458) = 20

E.g. 3—Solving Algebraically and Graphically

Solve the equation

e3 – 2x = 4

algebraically and graphically

E.g. 3—Solving Algebraically

The base of the exponential term is e.

So, we use natural logarithms to solve.

• You should check that this satisfies the original equation.

Solution 1

3 2

3 2

12

4

ln ln4

3 2 ln4

2 3 ln4

3 ln4 0.807

x

x

e

e

x

x

x

E.g. 3—Solving Graphically

We graph the equations y = e3–2x and y = 4

in the same viewing rectangle as shown.

• The solutions occur where the graphs intersect.

• Zooming in on the point of intersection, we see:

x ≈ 0.81

Solution 2

E.g. 4—Exponential Equation of Quadratic Type

Solve the equation e2x – ex – 6 = 0.

• To isolate the exponential term, we factor.

2

2(Law of Exponents)

(Zero-Product Property)

6 0

6 0

3 2 0

3 0 or 2 0

3 2

x x

x x

x x

x x

x x

e e

e e

e e

e e

e e

E.g. 4—Exponential Equation of Quadratic Type

The equation ex = 3 leads to x = ln 3.

However, the equation ex = –2 has no solution

because ex > 0 for all x.

• Thus, x = ln 3 ≈ 1.0986 is the only solution.

• You should check that this satisfies the original equation.

E.g. 5—Solving an Exponential Equation

Solve the equation 3xex + x2ex = 0.

• First, we factor the left side of the equation.

• Thus, the solutions are x = 0 and x = –3.

23 0

3 0

3 0

0 or 3 0

x x

x

xe x e

x x e

x x

x x

Logarithmic Equations

Logarithmic Equations

A logarithmic equation is one in which

a logarithm of the variable occurs.

• For example,

log2(x + 2) = 5

Solving Logarithmic Equations

To solve for x, we write the equation

in exponential form.

x + 2 = 25

x = 32 – 2

= 30

Solving Logarithmic Equations

Another way of looking at the first step

is to raise the base, 2, to each side.

2log2(x + 2) = 25

x + 2 = 25

x = 32 – 2 = 30

• The method used to solve this simple problem is typical.

Guidelines for Solving Logarithmic Equations

1. Isolate the logarithmic term on one side

of the equation.• You may first need to combine the logarithmic

terms.

2. Write the equation in exponential form

(or raise the base to each side).

3. Solve for the variable.

E.g. 6—Solving Logarithmic Equations

Solve each equation for x.

(a) ln x = 8

(b) log2(25 – x) = 3

E.g. 6—Solving Logarithmic Eqns.

ln x = 8

x = e8

Therefore, x = e8 ≈ 2981.

• We can also solve this problem another way:

Example (a)

ln 8

8

ln 8x

x

e e

x e

E.g. 6—Solving Logarithmic Eqns.

The first step is to rewrite the equation

in exponential form.

2

3

log 25 3

25 2

25 8

25 8

17

x

x

x

x

Example (b)

E.g. 7—Solving a Logarithmic Equation

Solve the equation

4 + 3 log(2x) = 16

• We first isolate the logarithmic term.

• This allows us to write the equation in exponential form.

E.g. 7—Solving a Logarithmic Equation

4

4 3 log 2 16

3 log 2 12

log 2 4

2 10

5000

x

x

x

x

x

E.g. 8—Solving Algebraically and Graphically

Solve the equation

log(x + 2) + log(x – 1) = 1

algebraically and graphically.

E.g. 8—Solving Algebraically

We first combine the logarithmic terms using

the Laws of Logarithms.

Solution 1

2

2

(Law 1)log 2 1 1

2 1 10

2 10

12 0

4 3 0

4 or 3

x x

x x

x x

x x

x x

x x

E.g. 8—Solving Algebraically

We check these potential solutions in

the original equation.

• We find that x = –4 is not a solution.

• This is because logarithms of negative numbers are undefined.

• x = 3 is a solution, though.

Solution 1

E.g. 8—Solving Graphically

We first move all terms to one side of

the equation:

log(x + 2) + log(x – 1) – 1 = 0

Solution 2

E.g. 8—Solving Graphically

Then, we graph

y = log(x + 2) + log(x – 1) – 1

• The solutions are the x-intercepts.

• So, the only solution is x ≈ 3.

Solution 2

E.g. 9—Solving a Logarithmic Equation Graphically

Solve the equation

x2 = 2 ln(x + 2)

• We first move all terms to one side of the equation

x2 – 2 ln(x + 2) = 0

E.g. 9—Solving a Logarithmic Equation Graphically

Then, we graph

y = x2 – 2 ln(x + 2)

• The solutions are the x-intercepts.

• Zooming in on them, we see that there are two solutions:

x ≈ 0.71 x ≈ 1.60

Application

Logarithmic equations are used

in determining the amount of light that

reaches various depths in a lake.

• This information helps biologists determine the types of life a lake can support.

Application

As light passes through water (or other

transparent materials such as glass or

plastic), some of the light is absorbed.

• It’s easy to see that, the murkier the water, the more light is absorbed.

Application

The exact relationship between light

absorption and the distance light travels

in a material is described in the next

example.

E.g. 10—Transparency of a Lake

Let:

• I0 and I denote the intensity of light before and after going through a material.

• x be the distance (in feet) the light travels in the material.

E.g. 10—Transparency of a Lake

Then, according to the Beer-Lambert Law,

where k is a constant depending on

the type of material.

0

1ln x

k

I

I

E.g. 10—Transparency of a Lake

(a) Solve the equation for I.

(b) For a certain lake k = 0.025 and the light

intensity is I0 = 14 lumens (lm).

Find the light intensity at a depth of 20 ft.

E.g. 10—Transparency of a Lake

We first isolate

the logarithmic term.

Example (a)

0

0

0

0

1ln

ln

kx

kx

xk

kx

e

e

I

I

I

I

I

I

I = I

E.g. 10—Transparency of a Lake

We find I using the formula from part (a).

• The light intensity at a depth of 20 ft is about 8.5 lm.

0

0.025 2014

8.49

kxe

e

I I

Example (b)

Compound Interest

Types of Interest

If a principal P is invested at an interest rate r

for a period of t years, the amount A of

the investment is given by:

Simple interest (for 1 year)

Interest compounded times per year

Interest compounded continuously

1

1nt

rt

n

A P r

rA t P

n

A t Pe

Compound Interest

We can use logarithms to determine

the time it takes for the principal

to increase to a given amount.

E.g. 11—Finding Term for an Investment to Double

A sum of $5000 is invested at an interest

rate of 5% per year.

Find the time required for the money

to double if the interest is compounded

according to the following method.(a) Semiannual

(b) Continuous

E.g. 11—Semiannually

We use the formula for compound interest

with

P = $5000, A(t) = $10,000, r = 0.05, n = 2

and solve the resulting exponential equation

for t.

Example (a)

E.g. 11—Semiannually

• The money will double in 14.04 years.

2

2

2

(Law 3)

0.055000 1 10000

2

1.025 2

log1.025 log2

2 log1.025 log2

log214.04

2log1.025

t

t

t

t

t

Example (a)

E.g. 11—Continuously

We use the formula for continuously

compounded interest with

P = $5000, A(t) = $10,000, r = 0.05

and solve the resulting exponential equation

for t.

Example (b)

E.g. 11—Continuously

• The money will double in 13.86 years.

0.05

0.05

0.05

5000 10,000

2

ln ln2

0.05 ln2

ln213.86

0.05

t

t

t

e

e

e

t

t

Example (b)

E.g. 12—Time Required to Grow an Investment

A sum of $1000 is invested at an interest rate

of 4% per year.

Find the time required for the amount to

grow to $4000 if interest is compounded

continuously.

E.g. 12—Time Required to Grow an Investment

We use the formula for continuously

compounded interest with

P = $1000, A(t) = $4000, r = 0.04

and solve the resulting exponential equation

for t.

E.g. 12—Time Required to Grow an Investment

• The amount will be $4000 in about 34 years and 8 months.

0.04

0.04

1000 4000

4

0.04 ln4

ln434.66

0.04

t

t

e

e

t

t

Annual Percentage Yield

If an investment earns compound

interest, the annual percentage yield

(APY) is:

• The simple interest rate that yields the same amount at the end of one year.

E.g. 13—Calculating the APY

Find the APY for an investment that earns

interest at a rate of 6% per year, compounded

daily.• After one year, a principal P will grow to:

• The formula for simple interest is: A = P(1 + r)

365

0.061 1.06183

365A P P

E.g. 13—Calculating the APY

Comparing, we see that:

1 + r = 1.06183

• Therefore, r = 0.06183.

• Thus, the annual percentage yield is 6.183%.