Solving Logarithmic Equations

By the end of this lesson, you will be able to solve logarithmic equations by converting to exponential form, using the one-to-one property, condensing logs before solving, and always checking for domain violations (extraneous solutions).

The Big Idea

Logarithmic equations require isolating the log expression and then converting to exponential form — or using the one-to-one property when logs with the same base appear on both sides. Crucially, logarithm arguments must be positive, so solutions that produce a zero or negative argument are extraneous and must be rejected.

Two Main Strategies

Strategy 1: Isolate the log, convert to exponential.

$$\log_b x = c \iff b^c = x$$

Strategy 2: Use the one-to-one property.

$$\log_b M = \log_b N \iff M = N \quad (M, N > 0)$$

Example 1 — Basic Conversion

Solve: $\log_4 x = 3$

Convert: $4^3 = x \implies x = 64$

Check: $\log_4 64 = 3$ ✓

Example 2 — Condense First

Solve: $\log x + \log(x - 3) = 1$

Condense (product property): $\log[x(x-3)] = 1$

Convert (base 10): $x(x-3) = 10^1 = 10$

$x^2 - 3x - 10 = 0 \implies (x-5)(x+2) = 0$

$x = 5$ or $x = -2$

Check $x = 5$: $\log 5 + \log 2 = \log 10 = 1$ ✓

Check $x = -2$: $\log(-2)$ undefined — extraneous.

Solution: $x = 5$

Example 3 — One-to-One Property

Solve: $\log_3(2x - 1) = \log_3(x + 4)$

Same base → same argument: $2x - 1 = x + 4 \implies x = 5$

Check: both arguments: $2(5)-1 = 9 > 0$ ✓; $5+4 = 9 > 0$ ✓

Example 4 — Natural Log Equation

Solve: $\ln(x + 2) + \ln(x - 1) = \ln 10$

Condense: $\ln[(x+2)(x-1)] = \ln 10$

One-to-one: $(x+2)(x-1) = 10$

$x^2 + x - 2 = 10 \implies x^2 + x - 12 = 0 \implies (x+4)(x-3) = 0$

$x = -4$ or $x = 3$

Check $x = -4$: $\ln(-2)$ undefined — extraneous.

Check $x = 3$: $\ln 5 + \ln 2 = \ln 10$ ✓

Solution: $x = 3$