Solve: $log_3 x + log_3 (x - 8) = 2$

Analyze the question: We have two logarithms with the same base 3 added together, and the sum equals 2. We need to find x.

Concept: Product rule: log_b M + log_b N = log_b (MN). Also, log_b y = k means y = b^k. Remember the domain: the logarithm needs positive arguments, so x > 0 and x - 8 > 0, which means x > 8.

Solution:

1. Combine the logs: log₃ [x(x - 8)] = 2

2. Convert to index form: x(x - 8) = 3² = 9

3. Expand and rearrange: x² - 8x - 9 = 0

4. Factorise (find two numbers with product -9 and sum -8: -9 and +1): (x - 9)(x + 1) = 0

5. So x = 9 or x = -1.

6. Check the domain: x must be greater than 8. x = -1 is rejected because log of a negative number is undefined. So x = 9.

Verification: log₃ 9 + log₃ (9 - 8) = log₃ 9 + log₃ 1 = 2 + 0 = 2 ✓

Answer: C (9)

Tip: Always check your roots in the original log equation. WASSCE often includes an extra root that makes a log argument negative, and options like 8 would make (x - 8) = 0, which is also invalid.