Question

If $3^{-m}\times \frac{1}{3^m}=\frac{1}{9}$, then find the value of $\frac{1}{24}\{{\left(3^m\right)}^2+{\left(\frac{1}{9^m}\right)}^{-1/2}\}$.

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{-1}{4}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}$

$3^{-m}\times \frac{1}{3^m}=\frac{1}{9}$

$3^{2m}=9$

$m=1$

Thus,

$\frac{1}{24}\left[\right(3^m{)}^2+\left(\frac{1}{9^m}{)}^{-1/2}\right]=\frac{1}{24}[3^2+(\frac{1}{9}{)}^{-1/2}]$

$=\frac{1}{24}[9+(\frac{1}{3}{)}^{-1}]$

$=\frac{1}{24}[9+3]=\frac{12}{24}=\frac{1}{2}$