Question

Find the value of n such that $\frac{4^{n+1}{2}^n-8^n}{2^{2n}}=\frac{3}{8}?$

Answer 1

$\frac{4^{n+1}{2}^n-8^n}{2^{2n}}=\frac{3}{8}$

On simplifying we get

$\frac{(2{)}^{2n+2}{2}^n-(2{)}^{3n}}{2^{2n}}=\frac{3}{8}$

$\frac{2^{3n}4-2^{3n}}{2^{2n}}=\frac{3}{8}$

$\frac{2^{3n}\left(3\right)}{2^{2n}}=\frac{3}{8}$

$2^n=\frac{1}{8}$

$(2{)}^n=(2{)}^{-3}$

On comparing powers we get

[n=-3]