Solve for x:

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

Given,

$\Rightarrow \sqrt{\dfrac{a}{b}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\dfrac{a}{b}\Big)^{\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x} \\[1em] \Rightarrow \Big(\dfrac{b}{a}\Big)^{-\dfrac{1}{2}} = \Big(\dfrac{b}{a}\Big)^{1 - 3x}$

Equating the exponents,

$\Rightarrow -\dfrac{1}{2} = 1 - 3x \\[1em] \Rightarrow -1 = 2(1 - 3x) \\[1em] \Rightarrow -1 = 2 - 6x \\[1em] \Rightarrow 6x = 2 + 1 \\[1em] \Rightarrow 6x = 3 \\[1em] \Rightarrow x = \dfrac{3}{6} = \dfrac{1}{2}.$

Hence, x = $\dfrac{1}{2}$.