Solving an Equation with the Laws of Logarithms

Log-law questions combine two skills: using $\log a + \log b = \log(ab)$ to collapse the equation, and, crucially, rejecting any root that a logarithm cannot accept. The rejection step is a mark, and it is the one most often dropped.

Solve $\log_3(x + 1) + \log_3(x - 1) = 1$. [5]

The working

Step 1, combine the logs with the addition law $\log_3 a + \log_3 b = \log_3(ab)$: $\log_3\big[(x + 1)(x - 1)\big] = 1 \quad \text{(M1)}$

Step 2, undo the log. $\log_3(\ \cdot\ ) = 1$ means the argument equals $3^1$: $(x + 1)(x - 1) = 3^1 = 3 \quad \text{(M1)}$

Step 3, solve the resulting equation: $x^2 - 1 = 3 \;\Rightarrow\; x^2 = 4 \;\Rightarrow\; x = \pm 2 \quad \text{(A1)}$

Step 4, reject invalid roots. Test each in the original logs. $x = -2$ gives $\log_3(-1)$ and $\log_3(-3)$, both undefined, so reject it. $x = 2$ gives $\log_3 3 + \log_3 1 = 1 + 0 = 1$ ✓: $x = 2 \quad \text{(M1, A1)}$

Where the marks are won and lost

• $\log_3(\cdot) = 1$ becomes $= 3^1$, not $= 1$ or $= 10$. The base is $3$; the “answer” of the log is the power.
• $(x+1)(x-1) = x^2 - 1$ (difference of two squares), a clean route to the quadratic.
• Rejecting $x = -2$ with a reason is the final mark. Giving "$x = \pm 2$" as the answer loses it because a negative argument breaks the original equation.

Common mistakes

• Writing $(x+1)(x-1) = 1$ (forgetting that $\log = 1$ means the argument is $3$, not $1$).
• Keeping both roots without checking validity.
• Multiplying the logs instead of adding their arguments.

Full method: Laws of Logarithms notes. Topic home: Logs & Exponentials pillar.