Exponential and Logarithmic Equations

By the end of this lesson, you will solve exponential and logarithmic equations at a pre-calculus level, including equations requiring substitution, systems, and checking domain restrictions.

Review: Core Techniques

Exponential equations: take the log of both sides, or use the same-base technique.

Logarithmic equations: condense to a single log, convert to exponential form, check domain.

Advanced Example 1 — Exponential System

Solve simultaneously: $e^x + e^y = 10$ and $e^x - e^y = 4$

Add: $2e^x = 14 \implies e^x = 7 \implies x = \ln 7$

Subtract: $2e^y = 6 \implies e^y = 3 \implies y = \ln 3$

Verify: $7 + 3 = 10$ ✓; $7 - 3 = 4$ ✓

Advanced Example 2 — Nested Logarithm

Solve: $\log_2(\log_3 x) = 2$

$\log_3 x = 2^2 = 4$

$x = 3^4 = 81$

Check: $\log_2(\log_3 81) = \log_2(4) = 2$ ✓

Advanced Example 3 — Equation Requiring Identity

Solve: $e^{2x} - 4e^x + 3 = 0$

Let $u = e^x$: $u^2 - 4u + 3 = (u-1)(u-3) = 0$

$e^x = 1 \implies x = 0$; $e^x = 3 \implies x = \ln 3$

Advanced Example 4 — Logarithm with Parameter

For what values of $k$ does $\ln x = kx$ have no solution?

$\ln x = kx \implies k = \frac{\ln x}{x}$

Let $h(x) = \frac{\ln x}{x}$ for $x > 0$. This function has a maximum at $x = e$: $h(e) = 1/e$.

If $k > 1/e$: no solution. If $k = 1/e$: exactly one solution. If $k \leq 0$ or $k < 1/e$: may have solutions.

Solving Logarithmic Inequalities

Solve: $\log_2(x-3) > 4$

$x - 3 > 2^4 = 16 \implies x > 19$

Also need $x - 3 > 0 \implies x > 3$ (already satisfied by $x > 19$).

Solution: $(19, \infty)$

Summary

Advanced techniques: substitution ($u = e^x$), systems of exponential equations, nested logs, parametric analysis.

Domain vigilance: log arguments must be positive — always check solutions in original.

Inequality: when $b > 1$, $\log_b$ is increasing so inequality direction preserved; when $0 < b < 1$, direction reverses.