Question

1. $log_{3}(x - 2) + log_{3}(x + 4) = 2$

Explanation:

Step1: Apply log product rule

$\log_3[(x-2)(x+4)] = 2$

Step2: Convert to exponential form

$(x-2)(x+4) = 3^2 = 9$

Step3: Expand left-hand side

$x^2 + 2x - 8 = 9$

Step4: Rearrange to quadratic equation

$x^2 + 2x - 17 = 0$

Step5: Solve quadratic formula

$x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-17)}}{2(1)} = \frac{-2 \pm \sqrt{72}}{2} = \frac{-2 \pm 6\sqrt{2}}{2} = -1 \pm 3\sqrt{2}$

Step6: Validate domain (logs need positive arguments)

For $x = -1 + 3\sqrt{2} \approx 3.24$, $x-2>0$ and $x+4>0$ (valid). For $x = -1 - 3\sqrt{2} \approx -5.24$, $x-2<0$ (invalid, discard).

Answer:

$x = -1 + 3\sqrt{2}$