An algebra problem by Aakhyat Singh

Let $\begin{cases} x = \log_b a & \implies a = b^x \\ y = \log_a b & \implies b = a^y \end{cases}$ $\implies a = b^x = a^{xy}$ $\implies xy = 1$ and $x = \dfrac 1y$ .

Now, we have:

$\begin{aligned} \frac {\log_a (\log_b a)}{\log_b (\log_a b)} & = \frac {\log_a x}{\log_b y} = \frac {\log_a \frac 1y}{\log_b y} = \frac {-\log_a y}{\log_b y} = - \frac {\log_a y}{\frac {\log_a y}{\log_a b}} = \boxed{-\log_a b} \end{aligned}$