Question

solve and answer the question​

Answer 1

$\begin{aligned}&\left[\Bigg\{\left(\frac{1}{7^2}\right)^{-2}\Bigg\}^{-1/3}\right]^{1/4}=7^m\\ \\ \Longrightarrow\ \ &\left[\Big\{\left(7^{-2}\right)^{-2}\Big\}^{-1/3}\right]^{1/4}=7^m\\ \\ \Longrightarrow\ \ &\left[\Big\{7^{(-2)\times (-2)}\Big\}^{-1/3}\right]^{1/4}=7^m\\ \\ \Longrightarrow\ \ &\left[\Big\{7^4\Big\}^{-1/3}\right]^{1/4}=7^m\\ \\ \Longrightarrow\ \ &\Big\{7^4\Big\}^{(-1/3)\times (1/4)}=7^m\end{aligned}$

$\Longrightarrow\ \ \Big\{7^4\Big\}^{-1/12}=7^m\\ \\ \\ \Longrightarrow\ \ &7^{-4/12}=7^m\\ \\ \\ \Longrightarrow\ \ 7^{-1/3}=7^m$

Therefore,

$\huge\boxed{\mathbf{m=-\dfrac{1}{3}}}$

Just used this identity.

$(a^m)^n=a^{mn}$

But here I've done the steps by doing the operations in braces after doing those in square brackets. And I've done those in parentheses first.

That's order of operation according to brackets. To solve any problem we have to do operations in parentheses first, then those in square brackets, and then in braces. But here we get same answer if we solve it without respect to the brackets.

Anyways,  -1/3  is the answer.