Question

If $\frac{4^{n+1}{.2}^n-8^n}{2^{2n}}=\frac{3}{8}$, then n = _____.

• A. $-1$
• B. $2$
• C. $-3$
• D. $-4$

Answer 1

The correct option is C $-3$

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

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

$\Rightarrow \frac{2^{3n}[2^2-1]}{2^{2n}}=\frac{3}{2^3}$

${32}^n={32}^{-3}$

$\Rightarrow n=-3$