Question

If ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m$ then find the value of m.

• A. $\frac{3}{2}$
• B. $-\frac{1}{7}$
• C. $\frac{2}{8}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is D

$-\frac{1}{3}$

${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m\Rightarrow {\left[{\left\{\right(7^{-2}{)}^{-2}\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m\Rightarrow {\left[\right(7^4{)}^{-\frac{1}{3}}]}^{\frac{1}{4}}=7^m\Rightarrow {\left(7^{-\frac{4}{3}}\right)}^{\frac{1}{4}}=7^m\Rightarrow {\left(7^{-\frac{4}{3}}\right)}^{\frac{1}{4}}=7^m\Rightarrow 7^{-1/3}=7^m\therefore m=-\frac{1}{3}$