Question

The value of $m$ for which ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=7^m$ is:

• A. a) $2$
• B. b) $-3$
• C. c) $-\frac{1}{3}$
• D. d) $\frac{1}{4}$

Answer 1

The correct option is C

c) $-\frac{1}{3}$

Explanation for correct options

Option (c): $-\frac{1}{3}$

We are given an equation ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=7^m$.

We need to solve for $m$.

Here, ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=7^m$

$$\begin{gathered} {\left[{\left\{{\left(7^2\right)}^2\right\}}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=7^m[\because {\left(\frac{1}{a}\right)}^{-m}=a^m] \\ {7}^{2\times 2\times -\frac{1}{3}\times \frac{1}{4}}=7^m[\because ({\left(a^m\right)}^n{)}^p=a^{mnp}] \\ {7}^{-\frac{1}{3}}=7^m[Onsimplifyingthepowers] \\ m=-\frac{1}{3}[\because If{a}^m=a^n,thenm=n] \end{gathered}$$

Hence, the value of $m$ is $-\frac{1}{3}$. Thus, the correct answer is option $\left(c\right)$.