Solving Exponential Equations

By the end of this lesson, you will be able to solve exponential equations by taking logarithms of both sides, using the same-base method, and applying the natural log for base-$e$ equations.

The Big Idea

Exponential equations have the variable in the exponent. Two strategies exist:

1. Same base: rewrite both sides with the same base, then set exponents equal.
2. Take the log: apply $\log$ or $\ln$ to both sides, then use the Power Property to bring the exponent down.

Strategy 2 is universal; Strategy 1 only works when both sides can be written with the same base.

Strategy 1 — Same Base

If $b^f = b^g$, then $f = g$.

Example: $2^{3x-1} = 2^{x+5}$

Same base ($2$), so: $3x - 1 = x + 5 \implies x = 3$

Requires: both sides expressed as powers of the same base.

$4^x = 8^{x-1}$: $4 = 2^2$, $8 = 2^3$ → $(2^2)^x = (2^3)^{x-1} \implies 2x = 3x - 3 \implies x = 3$

Strategy 2 — Taking Logarithms

Apply the log (or ln) to both sides, then use the Power Property:

$$a^x = b \implies x\log a = \log b \implies x = \frac{\log b}{\log a}$$

Example 1 — Basic Exponential Equation

Solve: $3^x = 50$

$$\log(3^x) = \log 50 \implies x\log 3 = \log 50 \implies x = \frac{\log 50}{\log 3} \approx \frac{1.699}{0.477} \approx 3.56$$

Example 2 — Natural Base

Solve: $5e^{2x} = 120$

Isolate the exponential: $e^{2x} = 24$

Apply $\ln$: $2x = \ln 24 \implies x = \dfrac{\ln 24}{2} \approx \dfrac{3.178}{2} \approx 1.589$

Example 3 — Quadratic-Type

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

Let $u = e^x$:

$u^2 - 5u + 6 = 0 \implies (u-2)(u-3) = 0$

$u = 2 \implies e^x = 2 \implies x = \ln 2 \approx 0.693$

$u = 3 \implies e^x = 3 \implies x = \ln 3 \approx 1.099$

Example 4 — Equation with Different Bases on Both Sides

Solve: $4^x = 7^{x-1}$

Take $\log$: $x\log 4 = (x-1)\log 7$

$x\log 4 = x\log 7 - \log 7$

$x(\log 4 - \log 7) = -\log 7$