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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.