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

Answer 1

Answer: (A)

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

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

$\Rightarrow {\left[\right(7^4{)}^{-\frac{1}{3}}]}^{\frac{1}{4}}=7^m$ [Since, $(a^m{)}^n=a^{mn}$]

$\Rightarrow {\left(7^{-\frac{4}{3}}\right)}^{\frac{1}{4}}=7^m$

$\Rightarrow 7^{(-\frac{4}{3}\times \frac{1}{4})}=7^m$

$\Rightarrow 7^{\frac{-1}{3}}=7^m$

Since, when the bases are equal across the equal sign then their exponents must be equal.

$\therefore m=-\frac{1}{3}$

Hence, Option A is correct.