di jawab y kawan #thank you

Penyelesaian:

Ingat: a^m . a^n = a^m+n
1/a^2 = a^-2
a^m = a^n , m = n

[tex] \frac{1}{4} . 2^(x^2 + 4x - 8) = \frac{1}{32} [/tex]
= [tex] \frac{1}{2^2} . 2^(x^2 + 4x - 8) = \frac{1}{32} [/tex]
= [tex]2^-2 . 2^(x^2 + 4x - 8) = 2^-5 [/tex]
= [tex]2^(x^2 + 4x - 10) = 2^-5 [/tex]
= x² + 4x - 10 = 5 
= x² + 4x - 5 = 0
(x + 5) (x - 1) = 0
x = -5 V x = 1

HP:{ x = -5 atau x = 1 }