Solve for x: log₂(x-1) + log₂(x+2) = 2 — JEE Mathematics

Domain constraints:

• $x - 1 > 0 \implies x > 1$
• $x + 2 > 0 \implies x > -2$
  Domain: $x > 1$.

Use the product rule for logarithms:

$$\log_2[(x-1)(x+2)] = 2$$

Convert from logarithmic to exponential form:

$$(x-1)(x+2) = 2^2$$

$$x^2 + x - 2 = 4$$

$$x^2 + x - 6 = 0$$

$$(x+3)(x-2) = 0 \implies x = -3 or x = 2$$

Since we require $x > 1$, we discard $x = -3$.
Thus, $x = 2$.

Answer: 2