dijawab y kawan kawan

Penyelesaian:

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

a. [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 }

b. [tex]7^(x^3 - 3x^2 - 10x - 2) = \frac{1}{49} [/tex]
= [tex]7^(x^3 - 3x^2 - 10x - 2) = 7^-2[/tex]
= [tex]x^3 - 3x^2 - 10x - 2 = -2[/tex]
= [tex]x^3 - 3x^2 - 10x = 0[/tex]
= [tex]x(x^2 - 3x - 10) = 0[/tex]
= [tex]x((x - 5) (x + 2)) = 0[/tex]
x = 0 ; x = 5 ; x = -2

HP:{ x = 0 , x = -2 , x = 5 }