Math Core

An exponential equation has the variable in an exponent, like $5^x = 40$ or $6e^{0.4x} = 45$ . These equations answer "when?" questions: when will an investment reach a goal, when will a population hit a target. There are two main strategies: rewrite both sides with the same base, or take a logarithm of both sides.

Strategy 1: Rewrite with a common base

Exponential functions are one-to-one: different inputs give different outputs. So if two powers of the same base are equal, their exponents must be equal.

Equal bases, equal exponents

For $b > 0$ and $b \ne 1$ : if $b^m = b^n$ , then $m = n$ .

This works whenever both sides can be written as powers of the same number.

Worked example: Common base

Solve each equation.

$4^{x + 1} = 8^x$
$9^x = \dfrac{1}{27}$

Solution.

Both $4$ and $8$ are powers of $2$ : $4 = 2^2$ and $8 = 2^3$ . Then

\left(2^2\right)^{x + 1} = \left(2^3\right)^x \;\Longrightarrow\; 2^{2x + 2} = 2^{3x} \;\Longrightarrow\; 2x + 2 = 3x \;\Longrightarrow\; x = 2.

Check: $4^3 = 64$ and $8^2 = 64$ . ✓

Write both sides as powers of $3$ : $9^x = 3^{2x}$ and $\dfrac{1}{27} = 3^{-3}$ . So $2x = -3$ and $x = -\dfrac{3}{2}$ .

Common mistake

Put parentheses around the whole exponent when you substitute. $\left(2^2\right)^{x + 1}$ is $2^{2x + 2}$ , not $2^{2x + 1}$ . The power rule multiplies $2$ by the entire exponent $x + 1$ .

Strategy 2: Take a logarithm of both sides

Most equations don't have a common base. There is no nice power of $5$ equal to $40$ . Instead, take the logarithm of both sides (any base works, but $\ln$ or $\log$ is what your calculator has), then use the power rule to bring the exponent down.

5^x = 40 \;\Longrightarrow\; \ln 5^x = \ln 40 \;\Longrightarrow\; x \ln 5 = \ln 40 \;\Longrightarrow\; x = \frac{\ln 40}{\ln 5} \approx 2.292.

You could also go straight to $x = \log_5 40$ by the definition of a logarithm and then use change of base. It's the same answer.

Solving by logarithms

Isolate the exponential expression (get $b^{\text{something}}$ alone on one side).
Take $\ln$ (or $\log$ ) of both sides.
Bring down the exponent with the power rule, then solve the resulting linear equation.

Worked example: Isolate first

Solve $3 \cdot 2^x + 4 = 70$ . Round to $3$ decimal places.

Solution. Isolate the power:

3 \cdot 2^x = 66 \;\Longrightarrow\; 2^x = 22.

Take $\ln$ of both sides:

x \ln 2 = \ln 22 \;\Longrightarrow\; x = \frac{\ln 22}{\ln 2} \approx \frac{3.091042}{0.693147} \approx 4.459.

That's reasonable, since $2^4 = 16$ and $2^5 = 32$ bracket $22$ .

Don't take the log before isolating. $\ln\left(3 \cdot 2^x + 4\right)$ can't be simplified, because there's no rule for the log of a sum.

Base e, and variables on both sides

When the base is $e$ , use $\ln$ , because $\ln e^{u} = u$ removes the base in one step.

Worked example: Natural base; two different bases

Solve each equation. Round to $3$ decimal places.

$6e^{0.4x} = 45$
$3^{x + 1} = 5^x$

Solution.

Divide by $6$ : $e^{0.4x} = 7.5$ . Take $\ln$ : $0.4x = \ln 7.5$ , so $x = \dfrac{\ln 7.5}{0.4} \approx \dfrac{2.014903}{0.4} \approx 5.037$ .
Take $\ln$ of both sides and bring down both exponents:

\begin{aligned} (x + 1)\ln 3 &= x \ln 5 \\ x\ln 3 + \ln 3 &= x \ln 5 \\ \ln 3 &= x\ln 5 - x \ln 3 \\ \ln 3 &= x(\ln 5 - \ln 3) \\ x &= \frac{\ln 3}{\ln 5 - \ln 3} \approx \frac{1.098612}{0.510826} \approx 2.151 \end{aligned}

Treat $\ln 3$ and $\ln 5$ as ordinary constants: gather the $x$ -terms on one side and factor out $x$ .

Equations in quadratic form

Some exponential equations are quadratics in disguise. Since $e^{2x} = \left(e^x\right)^2$ , substitute $u = e^x$ .

Worked example: Quadratic form

Solve $e^{2x} - 5e^x + 6 = 0$ .

Solution. Let $u = e^x$ . The equation becomes $u^2 - 5u + 6 = 0$ , which factors as $(u - 2)(u - 3) = 0$ . So $u = 2$ or $u = 3$ :

e^x = 2 \;\Longrightarrow\; x = \ln 2 \approx 0.693 \qquad e^x = 3 \;\Longrightarrow\; x = \ln 3 \approx 1.099.

If a factor had given $e^x = -4$ , you would reject it: $e^x$ is always positive.

Tip

Keep answers exact, like $\dfrac{\ln 22}{\ln 2}$ , until the last step, then round once. To check, substitute the rounded answer back in; you should land very close to the other side.