0,25^log(x-4) + 16^log(x+2) = 0

Logaritma

0,25'log (x - 4) + 16'log (x + 2) = 0
(4^-1)'log (x - 4) + (4²)'log (x + 2) = 0
4'log (x - 4)^-1 + 4'log √(x + 2) = 0
4'log (√(x + 2) /(x - 4)) = 4'log 1
√(x + 2) = x - 4
x + 2 = (x - 4)²
x + 2 = x² - 8x + 16
x² - 9x + 14 = 0
(x - 2)(x - 7) = 0
x = 2 atau x = 7

nilai x yg memenuhi :
x = 7