Find the value of x if [1÷64(x+2)]=[4(x?3)÷16x]

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

Answer Details

$[1÷{64}^{(x+2)}]=[4^{(x-3)}÷{16}^x]$
${64}^{-(x+2)}=\left[4^{(x-3)}\right]÷\left[{16}^x\right]$
Break down 4, 16 and 64 into smaller index numbers:
$2^{-6(x+2)}=2^{2(x-3)}÷2^{4\left(x\right)}$
$2^{-6x-12}=2^{2x-4x-6}$
$2^{-6x-12}=2^{-2x-6}$
− 6x − 12 = − 2x − 6
Collect the like terms:
−6x + 2x = −6 + 12
−4x =6
x = $\frac{6}{4}$
x = $\frac{-3}{2}$