Question

The calue of $m$ for which ${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m$, is

• A. $\frac{1}{4}$
• B. $-\frac{1}{3}$
• C. 2
• D. -3

Answer 1

The correct option is B $-\frac{1}{3}$

${\left[{\left\{{\left(\frac{1}{7^2}\right)}^{-2}\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m$
$\Rightarrow {\left[{\left\{\right(7^{-2}{)}^{-2}\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m$
$\Rightarrow {\left[{\left\{7^4\right\}}^{-\frac{1}{3}}\right]}^{\frac{1}{4}}=7^m$
$\Rightarrow 7^{\frac{-4}{3}\times \frac{1}{4}}=7^m$
$\Rightarrow 7^{\frac{-1}{3}}=7^m$
On comparing both the sides, we get
$m=\frac{-1}{3}$
Hence, (a) is correct answer.