8. solve the equation. $log_{4}(-x) + log_{4}(x + 10) = 2$ $x = square$ and $x = square$

Explanation:

Step1: Apply log product rule

$\log_4(-x(x+10)) = 2$

Step2: Convert to exponential form

$-x(x+10) = 4^2$

Step3: Simplify right-hand side

$-x(x+10) = 16$

Step4: Expand and rearrange to quadratic

$-x^2 -10x -16 = 0 \implies x^2 +10x +16 = 0$

Step5: Factor the quadratic

$(x+2)(x+8) = 0$

Step6: Solve for x, check domain

For $\log_4(-x)$ and $\log_4(x+10)$ to be defined: $-x>0 \implies x<0$, and $x+10>0 \implies x>-10$. Both $x=-2$ and $x=-8$ satisfy $-10<x<0$.

Answer:

$x=-2$ and $x=-8$