Superimposed the wrong way

Consider $\log _{ b }{ a }$ as x. then,

x^-1 = $\log _{ b }{ a }$ .

then, the simplification is,

$\frac { \log _{ a }{ x } }{ \log _{ b }{ { x }^{ -1 } } }$ .

= $\frac { \log _{ a }{ x } }{ -\log _{ b }{ { x } } }$ .

= $\frac { \log { x } }{ \log { a } } \times \frac { -\log { b } }{ \log { x } }$ .

After cancelling log(x) we get $-\log _{ a }{ b }$ .

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

I went about it the slow way, using extensive change of base. Reading the other solutions below, I now realize that this was totally unnecessary. Still, here you go: The expression in the problem is equal to the following: $= \dfrac{\dfrac{\log\left(\log_{b}a\right)}{\log a}}{\dfrac{\log\left(\log_{a}b\right)}{\log b}}$ $= \dfrac{\dfrac{\log\left(\dfrac{\log a}{\log b}\right)}{\log a}}{\dfrac{\log \left(\dfrac{\log b}{\log a}\right)}{\log b}}$ $= \dfrac{\dfrac{\log\left(\log a\right)-\log\left(\log b\right)}{\log a}}{\dfrac{\log\left(\log b\right) - \log\left(\log a\right)}{\log b}}$

Multiply by negative 1. $= \dfrac{\dfrac{\log\left(\log a\right)-\log\left(\log b\right)}{\log a}}{\dfrac{\log\left(\log a\right) - \log\left(\log b\right)}{\log b}}$ $= -\dfrac{\log b}{\log a}$ $= -\log_{a} b$

$\begin{aligned}\begin{aligned}\log_a{(\color{#3D99F6}{\log_ba})}=&\ \dfrac1{\log_{\color{#3D99F6}{\log_ba}}a}\\\log_b{(\color{#D61F06}{\log_ab})}=&\ \dfrac1{\log_{\color{#D61F06}{\log_ab}}b}\end{aligned}\end{aligned}$

Thus,

$\begin{aligned}\begin{aligned}\dfrac{\log_a{(\color{#3D99F6}{\log_ba})}}{\log_b{(\color{#D61F06}{\log_ab})}}=&\ \dfrac{\log_{\color{#D61F06}{\log_ab}}b}{\log_{\color{#3D99F6}{\log_ba}}a}\\=&\ \dfrac{\log_{(\color{#3D99F6}{\log_ba})^{-1}}b}{\log_{\color{#3D99F6}{\log_ba}}a}\\=&\ \dfrac{-\log_{\color{#3D99F6}{\log_ba}}b}{\log_{\color{#3D99F6}{\log_ba}}a}\\=&\ -\dfrac{\log{b}}{\log{a}}\\=&\ \boxed{-\log_ab}\end{aligned}\end{aligned}$