Question

If ${16}^{n+1}=64\times 4^{-n}$ the value of $n$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{9}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A $\frac{1}{3}$

${16}^{n+1}=64\times 4^{-n}$

$\Rightarrow (4^2{)}^{n+1}=4^3\times 4^{-n}$

Base is same on both sides of the equation so

$\Rightarrow 2n+2=3-n$

$\Rightarrow 2n+n=3-2$

$\Rightarrow 3n=1\Rightarrow n=\frac{1}{3}$