Solve the following equations:$ 4^{2 x}=(\sqrt[3]{16})^{-6 / y}=(\sqrt{8})^{2} $

Given:

\( 4^{2 x}=(\sqrt[3]{16})^{-6 / y}=(\sqrt{8})^{2} \)

To do: 

We have to solve the given equations.

Solution:

We know that,

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

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

$4^{2 x}=(\sqrt{8})^{2}$

$=8^{\frac{1}{2} \times 2}$

$=8$

$=(2)^{3}$

$\Rightarrow (2^{2})^{2 x}=2^{3}$

$\Rightarrow 2^{2 \times 2 x}=2^{3}$

$\Rightarrow 2^{4 x}=2^{3}$

Comparing both sides, we get,

$4 x=3$

$\Rightarrow x=\frac{3}{4}$

$(\sqrt[3]{16})^{-\frac{6}{y}}=(\sqrt{8})^{2}$

$\Rightarrow (\sqrt[3]{2^{4}})^{\frac{-6}{y}}=2^{3}$

$[(2)^{\frac{4}{3}\times\frac{-6}{y}}]=2^3$

Comparing both sides, we get,

$\frac{4}{3}\times\frac{-6}{y}=3$

$\frac{-8}{y}=3$

$y=\frac{-8}{3}$

The values of $x$ and $y$ are $\frac{3}{4}$ and $\frac{-8}{3}$ respectively.     

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Updated on: 10-Oct-2022

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