Exponential and Log Equations — AP® Precalculus

Part 1 · Hook

Why this matters

A medicine leaves the body so that

200 \cdot (0.8)^t

milligrams remain after

t

hours. When does it fall below

50

milligrams? The unknown is in the exponent, so ordinary algebra cannot reach it. Logarithms bring the exponent down, and that turns the question into one you can solve.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. Which is equal to $\log_2 x + \log_2 3$ for $x > 0$ ?

$\log_2(3x)$
$\log_2(x + 3)$
$3\log_2 x$
$\log_2 \frac{x}{3}$

Show the answer

Product property.

Correct: $\log_2(3x)$ :
$\log_2(x + 3)$ :
$3\log_2 x$ :
$\log_2 \frac{x}{3}$ :

2. $\ln$ and $e^x$ are inverse functions. What is $\ln(e^{5})$ ?

$5$
$e^5$
$5e$
$\ln 5$

Show the answer

$\ln$ undoes $e^x$ .

Correct: $5$ :
$e^5$ :
$5e$ :
$\ln 5$ :

3. Solve $x^2 - 2x - 8 = 0$ .

$x = 4$ or $x = -2$
$x = -4$ or $x = 2$
$x = 8$ or $x = -1$
$x = 2$ or $x = 4$

Show the answer

$(x - 4)(x + 2) = 0$ .

Correct: $x = 4$ or $x = -2$ :
$x = -4$ or $x = 2$ :
$x = 8$ or $x = -1$ :
$x = 2$ or $x = 4$ :

Part 4 · See it

See it first

Graph of y = 2 to the x and the dashed line y = 10, meeting at x = log base 2 of 10, about 3.32, just right of the point (3, 8). — The solution of $2^x = 10$ is where the graph of $y = 2^x$ meets $y = 10$ : $x = \log_2 10 \approx 3.32$ , between $3$ and $4$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The unknown is in an exponentisolate the power, then take a logarithm of both sides (or use a common base)
The unknown is inside logarithmscombine them into one logarithm, then rewrite in exponential form
Combining logarithms widens the domainsome roots of the new equation are extraneous: check each in the original
The equation is quadratic in $e^x$ substitute $u = e^x$ , solve, and reject any $u \le 0$
An inequality with $b^x$ or $\log_b x$ the direction stays for $b > 1$ and flips for $0 < b < 1$ ; respect the domain

Part 6 · Key ideas

Key ideas

Exponential equations: isolate $b^{(\ldots)}$ , then use a common base or take $\log$ or $\ln$ of both sides.
Logarithmic equations: get one logarithm on each side or one logarithm equal to a number, then rewrite as a power.
Check every answer in the original equation; a value that makes any logarithm's input $0$ or negative is extraneous.
Inverses of transformed functions: swap $x$ and $y$ , undo the outer steps in reverse order, and use the matching exponential or logarithm.

Part 7 · Misconception

A common mistake

The wrong idea: Solving $\ln x + \ln(x - 2) = \ln 8$ gives $x = 4$ or $x = -2$ , so both are answers.

What actually happens: $-2$ solves the quadratic $x^2 - 2x - 8 = 0$ but not the original equation, where $\ln(-2)$ is undefined. It is extraneous; the only solution is $x = 4$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $y = \log_2(x + 1)$ and the line $y = 3$

The curve is

f(x) = \log_2(x + 1)

. The dashed line is

y = 3

.

1. Solve $\log_2(x + 1) = 3$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Rewrite in exponential form: $x + 1 = 2^3 = 8$ , so $x = 7$ . The graph meets the line $y = 3$ at $(7, 3)$ .

Answer: 7

2. Solve $\log_2(x + 1) < 3$ .

$-1 < x < 7$
$x < 7$
$0 < x < 7$
$x > 7$

Show the answer

$f$ is increasing, so $f(x) < 3$ to the left of $x = 7$ . The domain needs $x + 1 > 0$ , so $x > -1$ . Together: $-1 < x < 7$ .

Correct: $-1 < x < 7$ : Below the line and inside the domain.
$x < 7$ : Inputs at or left of $x = -1$ are outside the domain, where $\log_2(x + 1)$ is undefined.
$0 < x < 7$ : For $-1 < x \le 0$ the graph is below the line too: $f(-0.5) = -1 < 3$ .
$x > 7$ : To the right of $7$ the graph is above the line.

Table

A student solves an equation

The equation is

\log_3 x + \log_3(x - 8)

= 2

.

A student solves log base 3 of x plus log base 3 of (x - 8) = 2, one step per row
Step	Work
1	$\log_3(x(x - 8)) = 2$
2	$x(x - 8) = 9$
3	$x^2 - 8x - 9 = 0$
4	$(x - 9)(x + 1) = 0$
5	$x = 9$ or $x = -1$

3. Which property of logarithms does step 1 use?

The product property
The quotient property
The power property
The change of base formula

Show the answer

A sum of two logarithms with the same base becomes the logarithm of a product.

Correct: The product property: $\log_3 x$ $+ \log_3(x - 8)$ $= \log_3(x(x - 8))$ .
The quotient property: The quotient property turns a difference into a quotient.
The power property: The power property moves an exponent; no exponent moves here.
The change of base formula: The base stays $3$ throughout.

4. Why does step 2 follow from step 1?

$\log_3 A = 2$ means $A = 3^2 = 9$
Both sides were squared
Both sides were multiplied by $3$ , and $2 \cdot 3 = 6$ was rounded to $9$
$\log_3$ was divided out of both sides

Show the answer

Rewriting $\log_3 A = 2$ in exponential form gives $A = 3^2 = 9$ .

Correct: $\log_3 A = 2$ means $A = 3^2 = 9$ : Logarithmic form to exponential form, with $A = x(x - 8)$ .
Both sides were squared: Nothing was squared; $9$ comes from $3^2$ .
Both sides were multiplied by $3$ , and $2 \cdot 3 = 6$ was rounded to $9$ : Multiplying by $3$ would give $6$ ; the $9$ is $3^2$ .
$\log_3$ was divided out of both sides: A logarithm is a function, not a factor; it is undone by its exponential.

5. Which conclusion about the student's answers is correct?

$x = 9$ is the solution; $x = -1$ is extraneous because $\log_3(-1)$ is undefined
Both $x = 9$ and $x = -1$ are solutions
$x = -1$ is the solution; $x = 9$ is extraneous because $9 - 8$ is small
Neither is a solution, because step 1 is not allowed

Show the answer

Check each in the original equation. $x = 9$ : $\log_3 9 + \log_3 1$ $= 2 + 0$ $= 2$ . $x = -1$ : $\log_3(-1)$ is undefined, so $-1$ is extraneous.

Correct: $x = 9$ is the solution; $x = -1$ is extraneous because $\log_3(-1)$ is undefined: $9$ checks, and $-1$ makes both logarithms undefined.
Both $x = 9$ and $x = -1$ are solutions: $-1$ solves the quadratic in step 3 but not the original equation.
$x = -1$ is the solution; $x = 9$ is extraneous because $9 - 8$ is small: $\log_3(9 - 8) = 0$ is allowed; $x = 9$ checks.
Neither is a solution, because step 1 is not allowed: Step 1 is fine for $x > 8$ ; it can add a false root, which is why we check.

6. Solve $\ln x + \ln(x - 2) = \ln 8$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Product property: $\ln(x(x - 2)) = \ln 8$ , so $x^2 - 2x - 8 = 0$ , $(x - 4)(x + 2) = 0$ . $x = -2$ makes $\ln x$ undefined, so it is extraneous. The solution is $x = 4$ .

Answer: 4

7. Solve $e^{2x} - e^x - 12 = 0$ . Select every solution.

$x = \ln 4$
$x = \ln 3$
$x = 4$
$x = -\ln 3$

Show the answer

Let $u = e^x$ : $u^2 - u - 12 = 0$ , $(u - 4)(u + 3) = 0$ . $e^x = 4$ gives $x = \ln 4$ . $e^x = -3$ is impossible, because $e^x > 0$ .

Correct: $x = \ln 4$ : $e^x = 4$ .
$x = \ln 3$ : The factor is $u + 3$ , which gives $e^x = -3$ , not $3$ ; and that has no solution.
$x = 4$ : $4$ is the value of $u = e^x$ ; then $x = \ln 4$ .
$x = -\ln 3$ : $e^{-\ln 3} = \frac{1}{3}$ , which does not solve $u^2 - u - 12 = 0$ .

Part 9 · Summary

Summary

To solve an exponential equation, isolate the power and use a common base or a logarithm. To solve a logarithmic equation, combine into one logarithm and rewrite as a power, then check for extraneous solutions. An equation quadratic in

e^x

is solved with

u = e^x

, rejecting roots where

e^x

would not be positive. Inequalities keep their direction for bases above

1

and flip for bases between

0

and

1

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of y=log2⁡(x+1)y = \log_2(x + 1) and the line y=3y = 3

A student solves an equation

Summary

What comes next

Connections

The graph of $y = \log_2(x + 1)$ and the line $y = 3$