Question

solve for x :

$\sqrt{a \b} = {b\a}^{1 - 3x}$
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Answer 1

$\sqrt{ \frac{a}{b} } = (\frac{b}{a} )^{1 - 3x}$

$\implies \sqrt{ \frac{a}{b} } = (\frac{a}{b} )^{3x - 1}$

$\implies \frac{a}{b}^{ \frac{1}{2} } = \frac{a}{b} ^{3x - 1}$

$\implies \frac{1}{2} = 3x - 1$

$\implies 3x = 1 + \frac{1}{2}$

$\implies 3x = \frac{3}{2}$

$\implies x = \frac{3}{2} \times \frac{1}{3}$

$\implies x = \frac{1}{2}$