Find the values of $x$ in each of the following:p>$ (\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32} $

Given:

\( (\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32} \)

To do: 

We have to find the value of $x$.

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,

$(\sqrt[3]{4})^{2 x+\frac{1}{2}}=\frac{1}{32}$

$\Rightarrow [(2^{2})^{\frac{1}{3}}]^{2 x+\frac{1}{2}}=\frac{1}{2^{5}}$

$\Rightarrow 2^{\frac{2}{3}(2 x+\frac{1}{2})}=2^{-5}$

Comparing both sides, we get,

$\Rightarrow \frac{2}{3}(2 x+\frac{1}{2})=-5$

$\Rightarrow \frac{4}{3} x+\frac{2}{6}=-5$

$\Rightarrow 8 x+2=-5 \times 6$

$\Rightarrow 8 x=-30-2$

$\Rightarrow x=\frac{-32}{8}$

$\Rightarrow x=-4$

The value of $x$ is $-4$.     

Tutorialspoint

Updated on: 10-Oct-2022

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