Solving Logarithmic Equations

Strategy for log equations

1. Combine logs using properties (product rule: $\log a + \log b = \log(ab)$).
2. Convert to exponential form ($\log_b x = y \iff b^y = x$).
3. Solve the resulting equation.
4. Check domain — you can only take the log of positive numbers!

Step-by-step: Solve $\log_2 x + \log_2(x - 2) = 3$

Step 1 — Combine using the product rule:

$\log_2[x(x - 2)] = 3$

Step 2 — Convert to exponential form ($\log_2(\text{stuff}) = 3$ means $\text{stuff} = 2^3$):

$x(x - 2) = 2^3 = 8$

Step 3 — Expand and solve the quadratic:

$x^2 - 2x = 8$ $x^2 - 2x - 8 = 0$ $(x - 4)(x + 2) = 0$ $x = 4 \quad \text{or} \quad x = -2$

Step 4 — Check domain. Both $\log_2(x)$ and $\log_2(x-2)$ require their arguments to be positive:

• $x = 4$: $\log_2(4) = 2$ ✓ and $\log_2(2) = 1$ ✓. Check: $2 + 1 = 3$ ✓
• $x = -2$: $\log_2(-2)$ is undefined ✗. Rejected!

Answer: $x = 4$ only.

Why checking the domain is essential

Logarithmic equations often produce extraneous solutions — values that satisfy the algebra but violate $\log(\text{positive only})$. Always plug answers back into the ORIGINAL equation.

Common mistakes

• Forgetting to check domain. $x = -2$ makes the algebra work but the log undefined.
• Splitting $\log(A \cdot B) = \log A + \log B$ the wrong way. This only works for multiplication inside the log, not addition.

Try it in the visualization

The graph shows $y = \log_2 x + \log_2(x-2)$ and $y = 3$. The intersection at $x = 4$ is the solution. The graph shows the function is undefined for $x \leq 2$, confirming $x = -2$ is impossible.