Interesting limit involving hordes of logarithms.

Question

$$\lim_{x\to 0^{+}}\ln (x\ln a)\ln \left(\dfrac{\ln ax}{\ln\frac xa}\right)=6$$ Find the value of $a$.

Answer: $e^3$

This struck me as an interesting problem and wanted to know if there are any more methods to solve it except the one I have used (written as an answer below).

Thanks!

Edit: for those requiring additional context, I have SOLVED the problem and my solution is posted below as an answer. I am merely looking for other ways to solve this question.

Answer 1

I shall make two substitutions which will be really useful. First, let $\ln a=b$. The left hand side of the equation becomes, $$(\ln x+\ln b)\ln\left(\dfrac{\ln x +b}{\ln x - b}\right)$$ Now, the second substitution: $\ln x=y$. Since $x\to 0^+\Rightarrow y\to -\infty$

So the limit now transforms to $$\lim_{y\to -\infty}(y+\ln b)\ln\left(1 + \dfrac{2b}{y-b}\right)$$

Using the limit

$$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1$$

We get the limit to be $$\lim_{y\to-\infty}\frac{2b(y+\ln b)}{y-b}=2b=6$$ Hence we have $b=3\Rightarrow a=e^3$.

Answer 2

$$\frac{\log ax}{\log\dfrac xa}=1+\frac{2\log a}{\log\dfrac xa}$$ so that

$$\log\frac{\log ax}{\log\dfrac xa}\sim\frac{2\log a}{\log\dfrac xa}.$$

As the denominator will simplify with $\log(x\log a))$, we are left with

$$2\log a=6.$$