Question

2*16^x-3*2^4x-1+7*4^2x-1=120

Answer 1

$2*16^x-3*2^{4x-1}+7*4^{2x-2}=120;$

$2*2^{4x}-3*2^{4x-1}+7*2^{4x-4}=120$

$2*2^{4x}-frac{3}{2}*2^{4x}+frac{7}{16}*2^{4x}=120;$

$2^{4x}(2-frac{3}{2}+frac{7}{16})=120;$

$frac{15}{16}2^{4x}=120;$

$2^{4x}=128=2^7;$

4x=7;

x=7/4;