Equations with logarithms are solved by turning them back into exponential equations with Switching Between Log and Exponential Form, often after combining logs with the Properties of Logarithms. The domain of the log is where the danger hides.
Key idea
Combine the logs into a single log on each side. Then either rewrite $\log_b(\text{stuff}) = c$ as $\text{stuff} = b^c$, or use $\log_b M = \log_b N \Rightarrow M = N$. Solve, and reject any answer that makes the input of an original log zero or negative.
Why it works
$\log_b x$ is only defined for $x > 0$. The product rule $\log_b x + \log_b(x - 2) = \log_b[x(x - 2)]$ is only true when both $x$ and $x - 2$ are positive, but $x(x - 2)$ can be positive when both are negative. So combining can let in answers that the original equation never allowed.
Worked examples
Worked example: One log
Solve $\log_3(2x + 1) = 2$.
- Exponential form: $2x + 1 = 3^2 = 9$.
- $x = 4$. Check: $2(4) + 1 = 9 > 0$.
Answer: $x = 4$
Worked example: A sum of logs
Solve $\log_2 x + \log_2(x - 2) = 3$.
- Product rule: $\log_2[x(x - 2)] = 3$, so $x(x - 2) = 8$.
- $x^2 - 2x - 8 = 0$, so $(x - 4)(x + 2) = 0$.
- $x = -2$ would need $\log_2(-2)$, which is undefined. Reject it.
Answer: $x = 4$
Worked example: A difference of logs
Solve $\ln(x + 6) - \ln x = \ln 3$.
- Quotient rule: $\ln\frac{x + 6}{x} = \ln 3$, so $\frac{x + 6}{x} = 3$.
- $x + 6 = 3x$, so $x = 3$. Both $3$ and $9$ are positive.
Answer: $x = 3$
Common mistake
Keeping a solution that makes the argument of a log negative. In the second example $x = -2$ satisfies $x(x - 2) = 8$, but $\log_2(-2)$ has no value. Always check each answer in the original equation, log by log.