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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Updated on: 10-Oct-2022

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