Question

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

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

Explanation:

Step1: Apply log product rule

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

Step2: Expand the product

$(x-2)(x+4) = x^2 + 2x - 8$

Step3: Rewrite in exponential form

$x^2 + 2x - 8 = 3^2 = 9$

Step4: Simplify to quadratic equation

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

Step5: Solve quadratic via formula

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

Step6: Check domain validity

Log arguments require $x-2>0$ and $x+4>0$, so $x>2$. $-1 - 3\sqrt{2} \approx -5.24$ is invalid.

Answer:

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