Math Core

A logarithmic equation has the variable inside a logarithm, like $\log_2(3x - 1) = 5$ . You'll see these when a formula is written with logs (the Richter and decibel scales, pH in chemistry) or when solving an exponential model backward. The good news: every method comes down to getting rid of the log. The catch: logs only accept positive inputs, so some "solutions" you find won't actually work.

Method 1: Rewrite in exponential form

When the equation has a single logarithm equal to a number, rewrite it as an exponential equation using the definition $\log_b x = y \iff b^y = x$ .

Worked example: One log equals a number

Solve each equation.

$\log_2(3x - 1) = 5$
$\ln(x + 2) = 3$

Solution.

In exponential form, $3x - 1 = 2^5 = 32$ . So $3x = 33$ and $x = 11$ . Check: $\log_2(33 - 1) = \log_2 32 = 5$ . ✓
The base of $\ln$ is $e$ , so $x + 2 = e^3$ and $x = e^3 - 2 \approx 20.085537 - 2 \approx 18.086$ .

Think of it as exponentiating both sides: raise the base to each side, and $b^{\log_b(\ldots)}$ collapses to what's inside.

Method 2: Set the arguments equal

Logarithmic functions are one-to-one, just like exponentials. If two logs with the same base are equal, their arguments are equal.

One-to-one property of logarithms

For $b > 0$ , $b \ne 1$ , and positive $M$ and $N$ : if $\log_b M = \log_b N$ , then $M = N$ .

Worked example: Log equals log

Solve $\log_5(2x + 3) = \log_5(x + 7)$ .

Solution. Set the arguments equal: $2x + 3 = x + 7$ , so $x = 4$ .

Check that both arguments are positive: $2(4) + 3 = 11$ and $4 + 7 = 11$ . ✓

Method 3: Condense first

When there are several logs, use the properties of logarithms to combine them into one, then use Method 1 or Method 2.

Worked example: Condense, then solve

Solve $\log_2 x + \log_2(x - 2) = 3$ .

Solution. Use the product property to condense the left side:

\log_2\left(x(x - 2)\right) = 3.

Rewrite in exponential form and solve the quadratic:

x(x - 2) = 2^3 \;\Longrightarrow\; x^2 - 2x - 8 = 0 \;\Longrightarrow\; (x - 4)(x + 2) = 0.

So $x = 4$ or $x = -2$ . Now check both in the original equation:

$x = 4$ : $\log_2 4 + \log_2 2 = 2 + 1 = 3$ . ✓
$x = -2$ : $\log_2(-2)$ is undefined. ✗

The only solution is $x = 4$ . The value $x = -2$ is an extraneous solution.

Why extraneous solutions appear

The original equation needs $x > 0$ and $x - 2 > 0$ . After condensing, the equation $\log_2\left(x(x - 2)\right) = 3$ only needs the product $x(x - 2)$ to be positive, which is also true when both factors are negative. Condensing widened the domain, and the extra solution slipped in through the gap.

Common mistake

Always check each solution in the original equation, not the condensed one. Every argument of every log must be positive. A negative solution isn't automatically wrong (for example, $x = -1$ works in $\log(x + 5) = \log 4$ ), and a positive one isn't automatically right. Substitute and look.

Worked example: A quotient

Solve $\log(x + 9) - \log x = 1$ .

Solution. Use the quotient property, then remember that $\log$ means base $10$ :

\log\frac{x + 9}{x} = 1 \;\Longrightarrow\; \frac{x + 9}{x} = 10^1 \;\Longrightarrow\; x + 9 = 10x \;\Longrightarrow\; x = 1.

Check: $\log 10 - \log 1 = 1 - 0 = 1$ . ✓

Tip

Before solving, write down the domain of the original equation. For $\log_2 x + \log_2(x - 2) = 3$ , the domain is $x > 2$ . Then any answer outside it, like $x = -2$ , can be crossed out immediately.