Logarithmic and Exponential Equations

Lessons

Senior 5 · Mathematics

Logarithmic and Exponential Equations

4 lessons

Lesson 2 of 4

Solving exponential and logarithmic equations

25 min

An unknown in the exponent

A sum of money growing at 6% a year is multiplied by ${1.06}^{n}$ after $n$ years. When has it doubled? We need ${1.06}^{n} = 2$ : the unknown sits in the exponent, where none of the methods for linear or quadratic equations can reach it. Logarithms bring it down: $n = \log_{1.06} 2 = \frac{\log 2}{\log 1.06} \approx 11.9$ years. This lesson collects the methods for equations with the unknown in an exponent or inside a logarithm, and the checks that keep false answers out.

Definition

Exponential equation: an equation in which the unknown appears in an exponent, such as $5^{x} = 12$ or $4^x - 5 \times 2^x + 4 = 0$ .

Logarithmic equation: an equation in which the unknown appears inside a logarithm, such as $\log_2 x + \log_2(x - 2) = 3$ .

Extraneous root: a value produced by the algebra that does not satisfy the original equation, usually because it lies outside the domain.

Example: $2^{x} = 8$ is exponential, with solution $x = 3$ ; $\log x = 2$ is logarithmic, with solution $x = 100$ .

Important

Three methods for exponential equations

Same base: write both sides as powers of one base; since $a^{x}$ is one-to-one, $a^{f(x)} = a^{g(x)} \iff f(x) = g(x)$ .
Take logarithms: if $a^{f(x)} = b$ with $b \gt 0$ , then $f(x) = \log_a b = \frac{\log b}{\log a}$ . There is no solution when $b \le 0$ .
Substitute: if the equation contains $a^{2x}$ and $a^{x}$ , put $t = a^x \gt 0$ and solve a quadratic in $t$ . Reject any $t \le 0$ .

Worked example 1: same base

Solve (a) $3^{2x-1} = 27^{x-2}$ (b) $2^{x^2 - 3x} = \frac{1}{4}$ .

Solution

$27^{x-2} = 3^{3x-6}$ , so $2 x - 1 = 3 x - 6$ and $x = 5$ .
$\frac{1}{4} = 2^{-2}$ , so $x^{2} - 3 x = - 2$ , $x^{2} - 3 x + 2 = 0$ , $(x - 1) (x - 2) = 0$ : $x = 1$ or $x = 2$ .

Worked example 2: taking logarithms

Solve, to 3 significant figures, (a) $5^{x} = 12$ (b) $2^{x+1} = 3^x$ .

Solution

$x = \log_5 12 = \frac{\log 12}{\log 5} = \frac{1.0792}{0.6990} = 1.54$ .
Take $\log$ of both sides: $(x + 1)\log 2 = x\log 3$ , so $x(\log 3 - \log 2) = \log 2$ and

x = \frac{\log 2}{\log 3 - \log 2} = \frac{\log 2}{\log 1.5} = \frac{0.30103}{0.17609} = 1.71.

Figure 1 shows the single crossing point.

Figure 1. The equation

2^{x+1} = 3^x

has exactly one solution, where the two graphs cross:

x = \frac{\log 2}{\log 1.5} \approx 1.71

Worked example 3: a hidden quadratic

Solve (a) $4^x - 5 \times 2^x + 4 = 0$ (b) $9^x - 3^{x+1} - 10 = 0$ .

Solution

$4^{x} = (2^{x})^{2}$ . With $t = 2^{x}$ : $t^{2} - 5 t + 4 = 0$ , $(t - 1) (t - 4) = 0$ . $2^{x} = 1$ gives $x = 0$ ; $2^{x} = 4$ gives $x = 2$ .
With $t = 3^{x}$ : $9^{x} = t^{2}$ and $3^{x+1} = 3t$ , so $t^{2} - 3 t - 10 = 0$ , $(t - 5) (t + 2) = 0$ . $t = - 2$ is impossible, since $3^x \gt 0$ . So $3^{x} = 5$ and $x = \frac{\log 5}{\log 3} \approx 1.46$ .

Worked example 4: an exponential inequality

Find the smallest whole number $n$ for which $0.8^n \lt 0.1$ .

Solution $\log$ is increasing, so taking $\log$ keeps the inequality: $n\log 0.8 \lt \log 0.1$ . Now $\log 0.8 \approx -0.0969$ is negative, so dividing by it reverses the sign:

n \gt \frac{\log 0.1}{\log 0.8} = \frac{-1}{-0.09691} = 10.32.

The smallest whole number is $n = 11$ . Check: $0.8^{10} \approx 0.107$ and $0.8^{11} \approx 0.086$ . ✓

Logarithmic equations

A logarithm exists only for a positive argument, so every logarithmic equation comes with conditions. Work in four steps:

Write the domain: every expression inside a logarithm must be positive.
Use the laws to reach one logarithm on each side, $\log_a A = \log_a B$ , or $\log_a A = c$ .
Remove the logarithms: $A = B$ , or $A = a^{c}$ , and solve.
Keep only the solutions that lie in the domain.

Figure 2.

\log_2 x + \log_2(x - 2)

is defined only for

x \gt 2

(shaded), so the algebra's second root

x = - 2

is not a solution.

Worked example 5: combine, solve, check

Solve (a) $\log_2 x + \log_2(x - 2) = 3$ (b) $\log(x + 3) - \log(x - 1) = \log 3$ .

Solution

Domain: $x \gt 0$ and $x \gt 2$ , so $x \gt 2$ . Then $\log_2[x(x - 2)] = 3$ , so $x^{2} - 2 x = 8$ , $(x - 4) (x + 2) = 0$ . $x = - 2$ is outside the domain, so the only solution is $x = 4$ (Figure 2). Check: $\log_2 4 + \log_2 2 = 2 + 1 = 3$ . ✓
Domain: $x \gt 1$ . Then $\log\frac{x + 3}{x - 1} = \log 3$ , so $x + 3 = 3 x - 3$ and $x = 3$ , which is in the domain.

Worked example 6: different bases, and a quadratic in $\log x$

Solve (a) $\log_3 x + \log_9 x = 6$ (b) $(\log x)^2 - 3\log x + 2 = 0$ .

Solution

Change base: $\log_9 x = \frac{\log_3 x}{\log_3 9} = \frac{1}{2}\log_3 x$ . So $\frac{3}{2}\log_3 x = 6$ , $\log_3 x = 4$ and $x = 81$ .
With $t = \log x$ : $t^{2} - 3 t + 2 = 0$ , so $t = 1$ or $t = 2$ : $x = 10$ or $x = 100$ . Both are positive, so both stand.

Worked example 7: a system

Solve $\log x + \log y = 3$ and $x - y = 15$ .

Solution Domain: $x \gt 0$ , $y \gt 0$ . The first equation gives $x y = 10^{3} = 1000$ . With $x = y + 15$ : $y^{2} + 15 y - 1000 = 0$ , so $y = \frac{-15 + \sqrt{225 + 4000}}{2} = \frac{-15 + 65}{2} = 25$ (the other root, $- 40$ , is negative). So $x = 40$ , $y = 25$ .

Warning

Check every root against the domain. In Worked example 5(a), $x = - 2$ satisfies $x^{2} - 2 x = 8$ but $\log_2(-2)$ does not exist. And when you divide an inequality by the logarithm of a number between 0 and 1, which is negative, reverse the inequality sign.

Tip

Choose the log that suits the calculator. To solve $a^{x} = b$ , use $x = \frac{\log b}{\log a}$ or $x = \frac{\ln b}{\ln a}$ : both give the same number, because the change-of-base formula works for any base.

bprp math basics: Solving an exponential equation with different bases. Watch how taking a logarithm of both sides turns 2^x = 5^(x+2) into a linear equation in x, and how the answer is tidied with the laws of logarithms.

✦

Practice

Give answers to 3 significant figures where they are not exact. The last two are harder.

Solve (a) $2^{3x} = 32^{x-1}$ (b) $25^x = \frac{1}{125}$ .
Solve $7^{x} = 40$ .
Solve $3^{2x-1} = 7$ .
Solve $7^x = 2 \times 3^x$ .
Solve $4^x + 2^{x+1} = 15$ .
Solve $\log_2(x + 1) + \log_2(x - 1) = 3$ .
Find the smallest whole number $n$ for which $1.05^n \gt 3$ .
Solve the system $2^x \times 4^y = 64$ , $\log_3 x - \log_3 y = 1$ .

Show answersHide answers

(a) $2^{3x} = 2^{5x-5}$ , so $x = \frac{5}{2}$ (b) $5^{2x} = 5^{-3}$ , so $x = -\frac{3}{2}$ .
$x = \frac{\log 40}{\log 7} = 1.90$ .
$2x - 1 = \frac{\log 7}{\log 3} = 1.771$ , so $x = 1.39$ .
$\left(\frac{7}{3}\right)^x = 2$ , so $x = \frac{\log 2}{\log\frac{7}{3}} = 0.818$ .
$t = 2^{x}$ : $t^{2} + 2 t - 15 = 0$ , $(t + 5) (t - 3) = 0$ ; $t = - 5$ is rejected, so $2^{x} = 3$ and $x = \frac{\log 3}{\log 2} = 1.58$ .
Domain $x \gt 1$ . $x^{2} - 1 = 8$ , so $x = 3$ ( $x = - 3$ is rejected).
$n \gt \frac{\log 3}{\log 1.05} = 22.5$ , so $n = 23$ .
$2^{x + 2y} = 2^6$ gives $x + 2 y = 6$ ; $\log_3\frac{x}{y} = 1$ gives $x = 3 y$ . So $5 y = 6$ : $y = \frac{6}{5}$ , $x = \frac{18}{5}$ , both positive.

Sources and credits

Written by Ganzaa: the explanations, the graphs, the worked examples, the practice items and the quiz questions.
Video: Solving an exponential equation with different bases, bprp math basics, embedded from YouTube.

1An unknown in the exponent

2Logarithmic equations

An unknown in the exponent

Logarithmic equations