TITLE: Simply this expression $\frac{\log_a(\log_b(a))}{\log_b(\log_a(b))}$
QUESTION [5 upvotes]: 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?

REPLY [2 votes]: 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.