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Solve the following equations:
\( \sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2 x} \), where \( a, b \) are distinct positive primes.
Given:
\( \sqrt{\frac{a}{b}}=\left(\frac{b}{a}\right)^{1-2 x} \), where \( a, b \) are distinct positive primes.
To do:
We have to solve the given equation.
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,
$\sqrt{\frac{a}{b}}=(\frac{b}{a})^{1-2 x}$
$\Rightarrow (\frac{a}{b})^{\frac{1}{2}}=(\frac{a}{b})^{-1+2 x}$
Comparing both sides, we get,
$\Rightarrow \frac{1}{2}=-1+2 x$
$\Rightarrow 2 x=1+\frac{1}{2}=\frac{3}{2}$
$\Rightarrow x=\frac{3}{2 \times 2}=\frac{3}{4}$
The values of $x$ is $\frac{3}{4}$.
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