Find the values of m and n if :

4^2m = $(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2$

Given,

4^2m = $(\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2$

Considering,

$\Rightarrow 4^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2m} = (\sqrt{8})^2 \\[1em] \Rightarrow 2^{4m} = 8 \\[1em] \Rightarrow 2^{4m} = 2^3 \\[1em] \Rightarrow 4m = 3 \\[1em] \Rightarrow m = \dfrac{3}{4}.$

Considering,

$\Rightarrow (\sqrt[3]{16})^{-\dfrac{6}{n}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\dfrac{1}{3} \times \Big(-\dfrac{6}{n}\Big)} = 8 \\[1em] \Rightarrow (16)^{-\dfrac{2}{n}} = 8 \\[1em] \Rightarrow (2^4)^{-\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{4 \times -\dfrac{2}{n}} = 2^3 \\[1em] \Rightarrow (2)^{-\dfrac{8}{n}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{n} = 3 \\[1em] \Rightarrow n = -\dfrac{8}{3}.$

Hence, $m = \dfrac{3}{4} \text{ and } n = -\dfrac{8}{3}$.