Question

If $\frac{(2^{-a}{)}^{-2}{8}^a+{32}^a}{2^{3b}4}=2^{-4}$, then 3b - 5 a =

Answer 1

$\frac{(2^{-a}{)}^{-2}\cdot (2^3{)}^a+(2^5{)}^a}{2^{3b}\cdot \left(2^2\right)}=2^{-4}$

$\frac{2^{2a}\cdot 2^{3a}+2^{5a}}{2^{3b}\cdot 2^2}=2^{-4}$

$\frac{2^{5a}+2^{5a}}{2^{2+3b}}$

$=\frac{{2.2}^{5a}}{2^{2+3b}}$

$=\frac{2^{5a+1}}{2^{2+3b}}$

$=2^{-4}$

$\Rightarrow 2^{(5a+1)-(2+3b)}=2^{-4}$

$\Rightarrow 5a-1-3b=-4$

$\Rightarrow 5a-3b=-3$

$\Rightarrow 3b-5a=3$.