2.13 - Exponential and Logarithmic Equations and Inequalities

Overview

This topic is about solving equations and inequalities where the variable is in an exponent or inside a log. The main tool is the inverse relationship: logs undo exponentials, and exponentials undo logs. You'll also check for extraneous solutions, rewrite exponentials in a different base, and build the inverse of a transformed exponential or log function. These skills show up all over the AP exam, especially in modeling questions.

The main tools

Inverse relationship: $\log_b\left(b^x\right) = x$ for every $x$ , and $b^{\log_b x} = x$ for $x > 0$ .

Switching forms: $\log_b c = a$ means the same thing as $b^a = c$ .

One-to-one: exponential and log functions are always increasing or always decreasing, so if $b^m = b^n$ then $m = n$ , and if $\log_b m = \log_b n$ (with $m, n > 0$ ) then $m = n$ .

Properties of exponents (Topic 2.4) and properties of logs (Topic 2.12) to combine or split terms.

Solving exponential equations

Strategy: get the exponential part by itself. Then either rewrite both sides with the same base, or take a log of both sides.

Example 1: Matching bases

Solve

9^x = 27^{x - 1}

Both 9 and 27 are powers of 3, so rewrite both sides with base 3:

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

3^{2x} = 3^{3x - 3}

The bases match, so the exponents must be equal:

2x = 3x - 3 \quad\Rightarrow\quad x = 3

Check:

9^3 = 729

and

27^2 = 729

. So

x = 3

Example 2: Isolate, then take a log

Solve

2 \cdot 5^{x - 1} + 3 = 19

Isolate the exponential part:

2 \cdot 5^{x - 1} = 16 \quad\Rightarrow\quad 5^{x - 1} = 8

Rewrite in log form (this is the same as taking

\log_5

of both sides):

x - 1 = \log_5 8

x = 1 + \log_5 8 = 1 + \frac{\ln 8}{\ln 5} \approx 2.292

Check:

2 \cdot 5^{1.292} + 3 \approx 2(8.00) + 3 = 19

Example 3: Different bases on both sides

Solve

2^{x + 1} = 5^x

The bases can't be matched, so take

\ln

of both sides and use the power property to bring the exponents down:

\ln\left(2^{x + 1}\right) = \ln\left(5^x\right)

(x + 1)\ln 2 = x\ln 5

Now

x

is in two places. Gather the

x

terms on one side and factor out

x

x\ln 2 + \ln 2 = x\ln 5

\ln 2 = x\ln 5 - x\ln 2 = x(\ln 5 - \ln 2)

x = \frac{\ln 2}{\ln 5 - \ln 2} \approx 0.756

Check:

2^{1.756} \approx 3.38

and

5^{0.756} \approx 3.38

Solving logarithmic equations

Strategy: use the properties from Topic 2.12 to combine everything into a single log, then rewrite in exponential form. Then check every answer in the original equation.

Example 4: Combine, then rewrite (AP-style)

Solve

\log_2(x + 2) + \log_2(x - 4) = 4

Use the product property to combine the left side:

\log_2\big((x + 2)(x - 4)\big) = 4

Rewrite in exponential form and solve the quadratic:

(x + 2)(x - 4) = 2^4 = 16

x^2 - 2x - 8 = 16

x^2 - 2x - 24 = 0

(x - 6)(x + 4) = 0 \quad\Rightarrow\quad x = 6 \ \text{ or } \ x = -4

Check both in the original equation. For

x = 6

\log_2 8 + \log_2 2 = 3 + 1 = 4

, so it works. For

x = -4

\log_2(-2)

is undefined, so

x = -4

is extraneous.

Answer:

x = 6

Extraneous solutions

An extraneous solution is a value you get from correct algebra that doesn't actually work in the original equation. Whether you solve with algebra or with a graph, you have to check your results against the limits of the math and of the context. With exponentials and logs, extraneous solutions come from three places:

Combining logs widens the domain. In Example 4, $\log_2(x + 2) + \log_2(x - 4)$ needs $x > 4$ , but $\log_2\big((x + 2)(x - 4)\big)$ also works for $x < -2$ . That's how $x = -4$ sneaked in.

Impossible exponential values. $b^x$ is always positive, so any step that leads to something like $e^x = -2$ gives no solution.

Context limits. A model might only make sense for $t \geq 0$ , or for whole numbers of people. A negative time can be fine mathematically and still be meaningless in the context.

Habit: before you solve a log equation, write down its domain (the input of every log must be positive). After you solve, cross out anything outside it.

Solving with a graph

You can also solve an equation by graphing each side as its own function and finding where the graphs intersect. For Example 4, graph

y_1 = \log_2(x + 2) + \log_2(x - 4)

and

y_2 = 4

(on a TI-84, use logBASE or type

\log_2(x + 2)

\frac{\ln(x + 2)}{\ln 2}

), then use the intersect tool. You'll only find

x = 6

, because

y_1

doesn't exist for

x \leq 4

. A graph of the original equation never shows answers outside its domain (the extraneous ones), so it's a good way to check your algebra. You still have to check its answers against the context.

Solving inequalities

Solve inequalities the same way as equations, with one extra question at every step: does this step keep the inequality pointing the same way?

If $b > 1$ , then $b^x$ and $\log_b x$ are increasing. Applying them to both sides keeps the direction: if $m < n$ , then $b^m < b^n$ .

If $0 < b < 1$ , they're decreasing, so applying them flips the direction.

Dividing both sides by a negative number flips it too. Watch out: $\ln b$ and $\log b$ are negative when $0 < b < 1$ .

For a log inequality, the answer also has to stay inside the domain.

Example 5: A base less than 1

Solve

\left(\frac{1}{2}\right)^x \leq 0.125

Take

\ln

of both sides. Since

\ln

is increasing, the direction stays the same:

x\ln(0.5) \leq \ln(0.125)

Now divide by

\ln(0.5)

, which is negative (about

-0.693

), so the inequality flips:

x \geq \frac{\ln 0.125}{\ln 0.5} = 3

Answer:

x \geq 3

. This makes sense:

\left(\frac{1}{2}\right)^x

is decreasing, so the outputs get smaller as

x

gets bigger.

\left(\frac{1}{2}\right)^3 = 0.125

, and every

x

past 3 gives something smaller still.

Example 6: A log inequality

Solve

\log_2(x - 3) < 4

Domain first:

x - 3 > 0

, so

x > 3

Since

2^x

is increasing, raise 2 to the power of each side without flipping:

x - 3 < 2^4 = 16 \quad\Rightarrow\quad x < 19

Answer: put both conditions together:

3 < x < 19

2.13 - Exponential and Logarithmic Equations and Inequalities

The graph shows it: the curve

y = \log_2(x - 3)

is below the line

y = 4

from the asymptote at

x = 3

up to

x = 19

. If you forgot the domain, you'd write

x < 19

, which wrongly includes values like

x = 0

where the log doesn't exist.

Example 7: An inequality in context (AP-style)

Suppose a town's population is modeled by

P(t) = 1200(1.05)^t

, where

t

is the number of years since 2020. When will the population be more than 2000?

1200(1.05)^t > 2000 \quad\Rightarrow\quad (1.05)^t > \frac{5}{3}

Take

\ln

of both sides, then divide by

\ln 1.05

. It's positive, so the direction stays:

t\ln 1.05 > \ln\left(\frac{5}{3}\right) \quad\Rightarrow\quad t > \frac{\ln(5/3)}{\ln 1.05} \approx 10.47

Answer: the population passes 2000 about 10.5 years after 2020, partway through 2030. (Check:

P(10) \approx 1955

and

P(11) \approx 2052

Rewriting exponentials in a different base

Logs let you rewrite an exponential with any other base

c

(with

c > 0

and

c \neq 1

b^x = c^{(\log_c b)x}

Why it works: by the inverse relationship,

b = c^{\log_c b}

. So

b^x = \left(c^{\log_c b}\right)^x = c^{(\log_c b)x}

This can reveal helpful information: it lets you compare functions with different bases on the same footing, and it lets you write a model in base

e

or base 10 when a problem asks for it.

$8^x = 2^{(\log_2 8)x} = 2^{3x}$ , so $8^x$ is $2^x$ with a horizontal dilation by a factor of $\frac{1}{3}$ .

$2^x = e^{(\ln 2)x} \approx e^{0.693x}$

$5^x = 10^{(\log 5)x} \approx 10^{0.699x}$

Example 8: Rewriting a model in base $e$

Write

f(t) = 500(1.06)^t

in the form

f(t) = 500e^{kt}

Use

c = e

, so

\log_c b = \ln 1.06

(1.06)^t = e^{(\ln 1.06)t}

f(t) = 500e^{(\ln 1.06)t} \approx 500e^{0.0583t}

Answer:

k = \ln 1.06 \approx 0.0583

. Both forms give the same outputs; for example,

f(10) \approx 895.4

either way (as long as you use the unrounded

k

Inverses of exponential and logarithmic functions

In Topic 2.10 you saw that

y = b^x

and

y = \log_b x

are inverses. Now the functions have additive transformations on them:

f(x) = ab^{(x + h)} + k \qquad \text{and} \qquad f(x) = a\log_b(x + h) + k

To find the inverse, reverse the mapping: list what

f

does to

x

, in order, then undo each step in the opposite order. (Swapping

x

and

y

and solving for

y

does the same thing.) For the exponential:

Step	What $f(x) = ab^{(x + h)} + k$ does to $x$	How the inverse undoes it
1	add $h$	subtract $h$ (last)
2	raise $b$ to that power	take $\log_b$
3	multiply by $a$	divide by $a$
4	add $k$	subtract $k$ (first)

So the inverse is

f^{-1}(x) = \log_b\left(\frac{x - k}{a}\right) - h

The same idea for the log: the inverse of

f(x) = a\log_b(x + h) + k

f^{-1}(x) = b^{(x - k)/a} - h

Don't memorize these formulas. Rebuild them by undoing the steps. Also remember that the domain and range swap: the horizontal asymptote

y = k

of the exponential becomes the vertical asymptote

x = k

of its inverse.

Example 9: Inverse of an exponential function

Find the inverse of

f(x) = 3 \cdot 2^{(x - 1)} + 5

and give its domain.

Write

y = f(x)

, swap

x

and

y

, and solve for

y

x = 3 \cdot 2^{(y - 1)} + 5

\frac{x - 5}{3} = 2^{(y - 1)}

\log_2\left(\frac{x - 5}{3}\right) = y - 1

f^{-1}(x) = \log_2\left(\frac{x - 5}{3}\right) + 1

Domain: the range of

f

y > 5

(its horizontal asymptote is

y = 5

), so the domain of

f^{-1}

x > 5

. The formula agrees:

\frac{x - 5}{3} > 0

means

x > 5

Check:

f(3) = 3 \cdot 2^2 + 5 = 17

, and

f^{-1}(17) = \log_2 4 + 1 = 3

Example 10: Inverse of a logarithmic function (AP-style)

Let

g(x) = 2\log_3(x + 4) - 1

. Find

g^{-1}(x)

and describe how the two graphs are related.

x = 2\log_3(y + 4) - 1

\frac{x + 1}{2} = \log_3(y + 4)

y + 4 = 3^{(x + 1)/2}

g^{-1}(x) = 3^{(x + 1)/2} - 4

Answer:

g^{-1}(x) = 3^{(x + 1)/2} - 4

. The graphs are reflections of each other over

y = x

, so every point swaps its coordinates. Since

g(5) = 2\log_3 9 - 1 = 3

, the point

(5, 3)

is on

g

and

(3, 5)

is on

g^{-1}

. Likewise

g(-1) = 1

, so

(-1, 1)

and

(1, -1)

match up. The vertical asymptote

x = -4

g

becomes the horizontal asymptote

y = -4

g^{-1}

Common mistakes

Taking a log of a sum term by term. $\ln\left(2^x + 3\right)$ is not $x\ln 2 + \ln 3$ . Isolate the exponential before taking a log.

Skipping the check. Log equations often produce extraneous solutions. Plug every answer back into the original equation.

Not flipping the inequality when you divide by $\ln b$ or $\log b$ with $0 < b < 1$ . Those logs are negative.

Forgetting the domain in a log inequality. $\log_2(x - 3) < 4$ is $3 < x < 19$ , not just $x < 19$ .

Undoing steps in the wrong order when building an inverse. Undo the last thing $f$ did first (usually subtract $k$ ).

Back Next

2.13 - Exponential and Logarithmic Equations and Inequalities

Overview

The main tools

Solving exponential equations

Example 1: Matching bases

Example 2: Isolate, then take a log

Example 3: Different bases on both sides

Solving logarithmic equations

Example 4: Combine, then rewrite (AP-style)

Extraneous solutions

Solving with a graph

Solving inequalities

Example 5: A base less than 1

Example 6: A log inequality

Example 7: An inequality in context (AP-style)

Rewriting exponentials in a different base

Example 8: Rewriting a model in base eee

Inverses of exponential and logarithmic functions

Example 9: Inverse of an exponential function

Example 10: Inverse of a logarithmic function (AP-style)

Common mistakes

Example 8: Rewriting a model in base $e$