Ta có
$\log_{\frac{\sqrt{b}}{a}} \left( \sqrt{\dfrac{a}{b}} \right) = \dfrac{1}{2} \log_{\frac{\sqrt{b}}{a}} \left( \dfrac{a}{b} \right)$
$= \dfrac{1}{2} \log_{\frac{\sqrt{b}}{a}} a -\dfrac{1}{2} \log_{\frac{\sqrt{b}}{a}} b$
$= \dfrac{1}{2} \dfrac{1}{\log_a \frac{\sqrt{b}}{a}} - \dfrac{1}{2} \dfrac{1}{\log_b \frac{\sqrt{b}}{a}}$
$= \dfrac{1}{2} \dfrac{1}{\log_a \sqrt{b} - \log_a a} - \dfrac{1}{2} \dfrac{1}{\log_b \sqrt{b} - \log_b a}$
$= \dfrac{1}{2} \dfrac{1}{\frac{1}{2} \log_a b - 1} - \dfrac{1}{2} \dfrac{1}{\frac{1}{2} - \log_b a}$
$= \dfrac{1}{\log_a b - 2} - \dfrac{1}{1 - 2\log_b a}$
Lại có $\log_a b = \sqrt{3}$, suy ra $\log_b a = \dfrac{1}{\sqrt{3}}$. Khi đó
$\log_{\frac{\sqrt{b}}{a}} \left( \sqrt{\dfrac{a}{b}} \right) = \dfrac{1}{\sqrt{3} - 2} - \dfrac{1}{1 - \frac{2}{\sqrt{3}}}$
$= \dfrac{\sqrt{3} + 2}{3-4} - \dfrac{\sqrt{3}}{\sqrt{3} - 2}$
$= -\sqrt{3} - 2 - \sqrt{3}(\sqrt{3} + 2)$
$= -\sqrt{3} - 2 - 3 - 2\sqrt{3} = -3\sqrt{3} - 5$.
