Stop Logging!

$\LARGE{Given}$ : $\LARGE{f(x)=log_{\frac{3}{2}}log_{\frac{1}{2}}log_{\pi}log_{\frac{\pi}{4}}.(x)}$ .

For "log" with base $\Large{\frac{3}{2}}$ to be defined :

$\Large{log_{\frac{1}{2}}log_{\pi}log_{\frac{\pi}{4}}.(x)>0}$

$\implies \Large{ 0 < log_{\pi}log_{\frac{\pi}{4}}.(x) <1}$

$\implies \Large{1< log_{\frac{\pi}{4}}.(x) < \pi}$

$\implies \Large{(\frac{\pi}{4})^{\pi} < (x) < \frac{\pi}{4}}$

CONCEPT HIDDEN ( usually mistaken) :

If we know $\Large{log_{a}.p > log_{a}.m}$

then if $\Large{ a>1 }$ , $\implies \Large{p > m}$

if $\Large{ 0<a<1}$ $\implies \Large{p < m}$ .

enjoy !

Since this is tagged with #JEE so I will use the favorite strategy of the exam takers of this prestigious exam. Let's see

the concept to be employed is for $\displaystyle\,log_a b$ to be defined we must have the following:

$(i)$ $a\neq1$

$(ii)$ $a>0$

$(iii)$ $b>0$

Now for the problem we start from the innermost logarithm which gives

$\displaystyle\,log_{\frac{\pi}{4}} x>0$

$\implies\,x<\left(\displaystyle\frac{\pi}{4}\right)^0$

This gives $\boxed{x<1}$ --- $(*)$

And $\displaystyle\,log_{\pi} log_{\frac{\pi}{4}} x >0$

$\implies\,log_{\frac{\pi}{4}} x >\pi$

$\implies\,x<\displaystyle\left(\frac{\pi}{4}\right)^{\pi}$ (Why ?? Since $\frac{\pi}{4}<1$ so inequality sign reverses)

Now the $\mathbf{strategy}$ part..Seeing all the options and checking which one of them satisfies $(*)$ , we luckily find that only one of them does so..So we can safely mark our answer without having to check for other nested logarithms as desired by a complete solution..

Use these 3 points,

(i) For $\log _{ a }{ b }$ to be defined, a & b should be greater than 0.

(ii) If $p<\log _{ a }{ b } <q$ & ${ a }>1$ then,

${ a }^{ p }<b<{ a }^{ q }$

(iii) If $p<\log _{ a }{ b } <q$ & ${ a }<1$ then,

${ a }^{ p }>b{ >a }^{ q }$