Find the value of p which satisfies the relation log2(p − 1) + 2 = log2(3p + 1).

Concept: For logarithms with the same base, the sum rule logb(a) + logb(c) = logb(ac) lets two log terms combine into one, and two equal logs of the same base force equal arguments: logb(x) = logb(y) ⇒ x = y (valid only when x, y > 0). A plain constant k can also join a log equation by rewriting it as k = logb(bk).

Applying this to the given equation:

1. Domain check: log2(p − 1) requires p − 1 > 0, i.e. p > 1.

2. Rewrite the constant 2 as a log2 term: 2 = log2(22) = log24, so the equation becomes log2(p − 1) + log24 = log2(3p + 1).

3. Combine the left side using the sum rule: log2[4(p − 1)] = log2(3p + 1).

4. Equal logs of the same base give equal arguments: 4(p − 1) = 3p + 1.

5. Expand and solve the resulting linear equation: 4p − 4 = 3p + 1 ⇒ p = 5.

6. Check the domain: p = 5 > 1 and 3p + 1 = 16 > 0 — both logarithm arguments stay valid.

Cross-check: substituting p = 5 back into the original relation — the left side is log2(5 − 1) + 2 = log24 + 2 = 2 + 2 = 4, and the right side is log2(3×5 + 1) = log216 = 4 — both sides equal 4, confirming p = 5.