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

Answer: (C)

$-3$

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

Let $2^n=t$

$\Rightarrow \frac{{4.4}^n{.2}^n-(2^3{)}^n}{(2^n{)}^2}=\frac{3}{8}$

$\Rightarrow \frac{4.t^3-t^3}{t^2}=\frac{3}{8}$

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

$\Rightarrow n=-3$