Simply this expression $\frac{\log_a(\log_b(a))}{\log_b(\log_a(b))}$

Question

How do I evaluate this $$\frac{\log_a(\log_b(a))}{\log_b(\log_a(b))}\tag{1}$$

My Efforts

I know this identity $$\log_a(b)=\frac{\log_d(b)}{\log_{d}(b)}\tag{2} $$

Let us fix a base say $e$ and I will call $\log$ to base $e$ simply as $\ln$

So let us see what numerator evaluates to

Using $(2)$ once we get,

$$\log_a(\log_b(a))=\frac{\ln(\log_b(a))}{\ln(a)}$$

Again using $(2)$ $$\log_a(\log_b(a))=\frac{\ln\left( \frac{\ln(a)}{\ln(b)}\right)}{\ln(a)}\tag{3}$$

Similarly denominator evaluates to $$\log_b(\log_a(b))=\frac{\ln\left( \frac{\ln(b)}{\ln(a)}\right)}{\ln(b)}\tag{4}$$

Using $(3)$ and $(4)$ in $(1)$, we get $$\log_a(b)=\frac{\log_d(b)}{\log_{d}(b)}\tag{5}= \frac{\frac{\ln\left( \frac{\ln(a)}{\ln(b)}\right)}{\ln(a)}}{\frac{\ln\left( \frac{\ln(b)}{\ln(a)}\right)}{\ln(b)}}$$

which further evaluates to $$-\frac{\ln(b)}{\ln(a)}$$ which is equal to $$-\log_a(b)$$

Am I correct?

Answer 1

Correct. A faster way of doing it without having to pass through the $\ln$ function is: \begin{equation} \frac{\log_a\log_ba}{\log_b\log_ab} \stackrel{(1)}{=} \frac{\log_a \frac{1}{\log_a(b)}}{\log_b\log_a b} \stackrel{(2)}{=} -\log_a b\frac{\log_a \log_a b}{\log_a \log_a b} = -\log_a b \end{equation} where we have used $\log_a(x)\log_b(a) = \log_b(x)$ in $(1)$ and $\log_b(x)\log_a(b) = \log_a(x)$ in $(2)$.

Answer 2

Your answer is correct except the fact,the original function is valid when $b\gt1 $ but in the resulting function $b \gt 0$so for equaling both of this,you must have to add the point that $b\gt 1$ from the first line.