Question

For a function $f:(e,\infty )\to R$ where$f\left(x\right)=\ln \left(\ln \right(\ln x\left)\right)$ the inverse can be obtained as

• A. $e^{e^{e^x}}$
• B. $e^{e^{e^{-x}}}$
• C. $e^{e^x}$
• D. Inverse doesn't exists for given function.

Answer 1

The correct option is A $e^{e^{e^x}}$

We have,
$f\left(x\right)=\ln \left(\ln \right(\ln x\left)\right)$
$f^{\prime }\left(x\right)=\frac{1}{\ln \left(\ln x\right)}\cdot \frac{1}{\ln x}\cdot \frac{1}{x}$
$\Rightarrow f^{\prime }\left(x\right)>0\forall x\in (e,\infty )$
So, $f\left(x\right)$ is strictly increasing function
$\therefore f\left(x\right)$ is one-one function
Also,
Range $=R$
Codomain $=R$
$\therefore f\left(x\right)$ is onto function
It is a bijection since it is both one-one and onto.
$\therefore$ Inverse exists.
Now,
$y=\ln \left(\ln \right(\ln x\left)\right)$
Replace $x$ by $y$,
$\Rightarrow x=\ln \left(\ln \right(\ln y\left)\right)$
$\Rightarrow \ln \left(\ln y\right)=e^x$
Continuing same way, we have $y=e^{e^{e^x}}$
Hence, $f^{-1}\left(x\right)=e^{e^{e^x}}$