Solving a Log Equation with Terms on Both Sides

The challenge here is the lone constant mixed in with logarithms. The fix: collect the logs using the subtraction law, then undo the log, remembering that "$\lg(\dots) = 1$" means the argument equals $10^1$.

Solve $\lg(x + 2) = 1 + \lg(x - 1)$. [5]

The working

Step 1, gather the logs on one side: $\lg(x + 2) - \lg(x - 1) = 1 \quad \text{(M1)}$

Step 2, combine with the subtraction law $\lg a - \lg b = \lg\frac{a}{b}$: $\lg\!\left(\frac{x + 2}{x - 1}\right) = 1 \quad \text{(M1)}$

Step 3, undo the log. Base $10$, so the argument equals $10^1 = 10$: $\frac{x + 2}{x - 1} = 10 \quad \text{(M1)}$

Step 4, solve: $x + 2 = 10(x - 1) \;\Rightarrow\; x + 2 = 10x - 10 \;\Rightarrow\; 12 = 9x \;\Rightarrow\; x = \frac{4}{3} \quad \text{(A1)}$

Check the arguments are positive: $x - 1 = \frac13 > 0$ and $x + 2 = \frac{10}{3} > 0$, so $x = \frac{4}{3}$ is valid. (A1)

Where the marks are won and lost

• The constant $1$ becomes an exponent: $\lg(\text{arg}) = 1$ means arg $= 10^1$, not $= 1$. This is the decisive step.
• Use the subtraction law to collect the two logs into one, don’t try to “cancel” the $\lg$s term by term.
• Check the arguments are positive. Here both are, so no rejection, but the check is expected and occasionally rules a solution out.

Common mistakes

• Writing $\frac{x+2}{x-1} = 1$ (forgetting the constant makes it $10$, not $1$).
• Mishandling the subtraction law, or dividing the logs instead of subtracting their arguments.
• Skipping the positive-argument check.

Full method: Solving Log & Exponential Equations notes. Topic home: Logs & Exponentials pillar.