Find the value of x if \( [ 1 ÷ 64^{(x + 2)}]= [4^{(x − 3)} ÷ 16^x ] \)

A.

\( \frac{3}{2} \)

B.

\( \frac{2}{3} \)

C.

\( \frac{1}{3} \)

D.

\( -\frac{3}{2} \)

Correct answer is D

\( [ 1 ÷ 64^{(x + 2)}]= [4^{(x − 3)} ÷ 16^x ] \)

\( 64^{−(x + 2)} = [4^{(x − 3)}] ÷[16^x] \)

breakdown 4,16,64 into a small index no

\( 2^{−6(x + 2)} = 2^{2(x − 3)} ÷ 2^{4(x)} \)

\( 2^{−6x− 12} = 2^{2x − 4x − 6} \)

\( 2^{−6x −12} = 2^{−2x − 6} \)

− 6x − 12 = − 2x − 6

Collect the like term

−6x + 2x = −6 + 12

−4x =6

x = \( \frac{6}{4} \)

x = \( \frac{−3}{2} \)