Free AP Precalculus Exponential and Logarithmic Functions

AP Precalculus · Exponential and Logarithmic Functions · free lesson

Exponential and Logarithmic Equations and Inequalities

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

Undoing an exponential: the logarithm

To solve an equation you undo what has been done to the unknown, in the reverse order. An exponential is undone by a logarithm of the same base and a logarithm by an exponential, because they are inverse functions: $\log_b \left(b^x\right) = x$ and $b^{\log_b x} = x$ for $x > 0$ . So the key step is always one of these:

Rule.

For $b > 0$ , $b \ne 1$ : $b^{x} = c$ (with $c > 0$ ) exactly when $x = \log_b c$ ; and $\log_b x = k$ exactly when $x = b^{k}$ . The equation $b^{x} = c$ has no solution when $c \le 0$ .

When both sides can be written as powers of one base, set the exponents equal, as in the topic on rewriting exponentials. Otherwise isolate the power first, then take the logarithm. For $5 \cdot 2^{x - 1} + 3 = 83$ :

\begin{aligned} 5 \cdot 2^{x - 1} &= 80 \\ 2^{x - 1} &= 16 \\ x - 1 &= \log_2 16 = {4} \\ x &= 5 \end{aligned}

For $3^{x} = 20$ there is no common base. Take $\log_3$ of both sides:

$x = \log_{3} 20$ , an exact answer, which is about $2.727$ . Leave it as a logarithm unless a decimal is asked for: on the no-calculator part of the AP exam, $x = \log_3 20$ (or $\frac{\ln 20}{\ln 3}$ ) is the answer. Move the slider: the line $y = c$ meets $y = 3^{x}$ once for every $c > 0$ , at $x = \log_3 c$ , and never for $c \le 0$ .

The same steps with base $e$ use $\ln$ : $e^{2x} = 7$ gives $2x = \ln 7$ . And any base can be rewritten with base $e$ , since $b = e^{\ln b}$ : $3^x = \left(e^{\ln 3}\right)^x = e^{(\ln 3) x}$ .

Logarithmic equations, and answers to throw out

To solve $\log_3 x + \log_3 (x - 8) = 2$ , combine the logarithms with the product rule, change to exponential form and solve:

\begin{aligned} \log_3 \left(x(x - 8)\right) &= 2 \\ x(x - 8) &= 3^{2} = 9 \\ x^2 - 8x - 9 &= 0 \end{aligned}

The quadratic has the solutions $9$ and $-1$ . But the original equation needs $x > 0$ and $x - 8 > 0$ , so $x = {-1}$ is not in its domain: $\log_3 ({-1})$ is not defined. The only solution is $x = 9$ .

The picture shows why the extra answer appears. Combining the logarithms gave $\log_3 \left(x(x - 8)\right)$ , which is also defined for $x < 0$ (red); the original left side (blue) is defined only for $x > 8$ . The line $y = 2$ meets the red curve twice and the blue one once.

Equations quadratic in $b^{x}$

In $4^x - 3 \cdot 2^x - 4 = 0$ , write $4^x = \left(2^x\right)^2$ and let $u = 2^{x}$ :

\begin{aligned} u^2 - 3u - 4 &= 0 \end{aligned}

gives $u = {4}$ or $u = {-1}$ . Now undo the substitution: $2^{x} = 4$ gives $x = 2$ , while $2^{x} = -1$ has no solution, because $2^{x} > 0$ for every $x$ . So $x = 2$ is the only answer.

Inequalities: the base decides the direction

For $b > 1$ , $b^{x}$ and $\log_b x$ are increasing, so applying them keeps an inequality's direction. For $0 < b < 1$ they are decreasing, so the direction turns:

Inequality	Step	Solution
$2^{x} > 8$	$x > \log_2 8$ , direction kept	${x > 3}$
$\left(\frac{1}{2}\right)^x > 8$	$x < \log_{1/2} 8$ , direction turned	${x < -3}$
$\log (x - 1) < 1$	$x - 1 < 10^{1}$ , and $x - 1 > 0$ for the domain	${1 < x < 11}$

The picture shows the second: $\left(\frac{1}{2}\right)^x$ is above the line $y = 8$ exactly to the left of the crossing, where $x < -3$ , the side of the green line $x = -3$ that is shaded.

Inverses of transformed exponential and logarithmic functions

To find the inverse of $f(x) = 3 \cdot 2^{x - 1} + 4$ , write $y = f(x)$ , swap $x$ and $y$ , and undo the steps in reverse order:

\begin{aligned} x &= 3 \cdot 2^{y - 1} + 4 \\ \frac{x - 4}{3} &= 2^{y - 1} \\ y - 1 &= \log_2 \frac{x - 4}{3} \end{aligned}

So $f^{-1}(x) = \log_2 \frac{x - 4}{3} + 1$ ; the engine gives $\log_{2}(\frac{1}{3}(x - 4)) + 1$ , with domain ${x > 4}$ . The domain of $f^{-1}$ is the range of $f$ : the horizontal asymptote $y = 4$ of $f$ becomes the vertical asymptote $x = 4$ of $f^{-1}$ . Click the picture to watch the two graphs drawn, mirror images over $y = x$ .

Warning.

The usual slips: keeping an answer that makes a logarithm's argument zero or negative; accepting $b^{x} = -1$ as if it had a solution; reading $\log a - \log b$ as $\frac{\log a}{\log b}$ (it is $\log \frac{a}{b}$ ); forgetting to turn an inequality for a base between 0 and 1; and answering $e^{x} = 5$ with $\log 5$ , the base-10 logarithm, instead of $\ln 5$ .

Note.

On the AP exam's symbolic question (no calculator) show each step, give exact answers such as $\log_3 20$ or $\frac{\ln 7}{2}$ , and say why you drop a candidate: " $x = {-1}$ is rejected because $\log_3 ({-1})$ is not defined."

Worked examples

Each example is one kind of equation or inequality the AP exam asks about. Every number in the working is found by the engine. In each one: isolate the power or the logarithm, undo it with its inverse, and check the answer against the domain.

Example.

1. A common base, and no common base (no calculator). Solve (a) $\left(\frac{1}{8}\right)^x = 4^{x + 5}$ and (b) $4e^{3x} - 1 = 19$ , giving exact answers.

(a) Both bases are powers of 2: $\frac{1}{8} = 2^{-3}$ and $4 = 2^{2}$ . Then set the exponents equal:

\begin{aligned} 2^{-3x} &= 2^{2x + 10} \\ -3x &= 2x + 10 \\ x &= -2 \end{aligned}

(b) Here there is no common base. Isolate the power of $e$ , then take $\ln$ of both sides:

\begin{aligned} 4e^{3x} &= 20 \\ e^{3x} &= 5 \\ 3x &= \ln 5 \\ x &= \frac{1}{3} \times \ln(5) \end{aligned}

That is $x = \frac{\ln 5}{3}$ , about $0.536$ . Answering $\frac{\log 5}{3}$ uses base 10 where the base is $e$ .

Result.

(a) $x = -2$ . (b) $x = \frac{\ln 5}{3}$ .

Example.

2. An exact answer in logarithms (no calculator). Solve $7^{x + 2} = 3$ . Then rewrite $7^{x}$ as an exponential with base $e$ .

Take $\log_7$ of both sides; $\log_7 \left(7^{x + 2}\right) = x + 2$ because the logarithm undoes the exponential:

\begin{aligned} x + 2 &= \log_7 3 \\ x &= -2 + \log_{7} 3 \end{aligned}

So $x = \log_7 3 - 2$ , about $-1.435$ . A quick check of size: $7^{0} = 1 < 3 < 7^{1}$ , so $x + 2 = \log_7 3$ is between 0 and 1, and $x$ is between $-2$ and $-1$ .

For the rewrite, $7 = e^{\ln 7}$ , so by the power rule $7^x = \left(e^{\ln 7}\right)^x = e^{(\ln 7)x}$ : an exponential with base $e$ and the constant $\ln 7$ in the exponent.

Result.

$x = \log_7 3 - 2$ , and $7^x = e^{(\ln 7)x}$ .

Example.

3. A logarithmic equation with an answer to reject (no calculator). Solve $\log_3 x + \log_3 (x + 8) = 2$ .

The equation is defined only where both arguments are positive: $x > 0$ (and then $x + 8 > 0$ too). Combine with the product rule, then change to exponential form:

\begin{aligned} \log_3 \left(x(x + 8)\right) &= 2 \\ x(x + 8) &= 3^{2} \\ x^2 + 8x - 9 &= 0 \end{aligned}

The quadratic gives $1$ and $-9$ . The candidate $x = -9$ is rejected because $\log_3 ({-9})$ is not defined. Check $x = 1$ : $\log_3 1 + \log_3 9 = 0 + 2 = 2$ .

Result.

$x = 1$ only; $x = -9$ is outside the domain.

Example.

4. An equation quadratic in $e^{x}$ (no calculator). Solve $e^{2x} - 5e^x + 6 = 0$ . Then solve $9^x - 2 \cdot 3^x - 3 = 0$ .

Since $e^{2x} = \left(e^x\right)^2$ , let $u = e^{x}$ . Then $u^2 - 5u + 6 = 0$ , so $u = {3}$ or $u = {2}$ . Both are positive, so both give an answer: $e^{x} = 2$ gives $x = \ln 2$ and $e^{x} = 3$ gives $x = \ln 3$ . The engine agrees: $x = {\ln(2)}$ or $x = {\ln(3)}$ .

In the second, $9^x = \left(3^x\right)^2$ ; let $v = 3^{x}$ . Then $v^2 - 2v - 3 = 0$ , so $v = {3}$ or $v = {-1}$ . Now $3^{x} = -1$ has no solution, since $3^{x} > 0$ for every $x$ ; $3^{x} = 3$ gives $x = 1$ .

Result.

$x = \ln 2$ or $x = \ln 3$ ; and $x = 1$ for the second equation.

Example.

5. An inequality in context (calculator). A patient is given 200 milligrams of a medicine. The amount left in the body $t$ hours later is $A(t) = 200(0.8)^t$ milligrams. For which times is less than 50 milligrams left? Then solve $\log_2 (x - 3) \ge 4$ .

Divide by 200: $0.8^{t} < 0.25$ . The base 0.8 is between 0 and 1, so $0.8^{t}$ is decreasing, and taking $\log_{0.8}$ of both sides turns the inequality:

\begin{aligned} t &= \log_{0.8} 0.25 \approx {6.213} \end{aligned}

So less than 50 milligrams is left from about $6.213$ hours after the dose on. Check with a time on each side: $A(6) \approx {52.429}$ milligrams (more than 50) and $A(7) \approx {41.943}$ milligrams (less).

For $\log_2 (x - 3) \ge 4$ : the base 2 is greater than 1, so the direction stays. $x - 3 \ge 2^4 = 16$ gives $x \ge 19$ , and those $x$ are inside the domain $x > 3$ . The engine gives ${x \ge 19}$ .

Result.

Less than 50 milligrams is left for $t > \log_{0.8} 0.25$ , about $6.213$ hours; and $\log_2 (x - 3) \ge 4$ for ${x \ge 19}$ .

Example.

6. The inverse of a transformed exponential (no calculator). Find the inverse of $f(x) = 2e^{x - 1} + 3$ and its domain.

Write $x = 2e^{y - 1} + 3$ and undo the steps in reverse order: subtract 3, divide by 2, take $\ln$ , add 1.

\begin{aligned} x - 3 &= 2e^{y - 1} \\ \frac{x - 3}{2} &= e^{y - 1} \\ y - 1 &= \ln \frac{x - 3}{2} \end{aligned}

So $f^{-1}(x) = \ln \frac{x - 3}{2} + 1$ ; the engine gives $\ln(\frac{1}{2}(x - 3)) + 1$ with domain ${x > 3}$ , which is the range of $f$ (the graph of $f$ lies above its asymptote $y = 3$ ). Check: $f(1) = 2e^0 + 3 = 5$ , and $f^{-1}(5) = \ln 1 + 1 = {1}$ .

Result.

$f^{-1}(x) = \ln \frac{x - 3}{2} + 1$ , for ${x > 3}$ .

Practice

Answer AP-style questions on exponential and logarithmic equations and inequalities. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise exponential and logarithmic equations and inequalities with new numbers

Exponential and Logarithmic Equations and Inequalities

The idea

Undoing an exponential: the logarithm

Logarithmic equations, and answers to throw out

Equations quadratic in bxb^{x}bx

Inequalities: the base decides the direction

Inverses of transformed exponential and logarithmic functions

Worked examples

Worked examples

Practice

Equations quadratic in $b^{x}$