# How do you find the inverse of f(x)=ln(2+ln(x))?

Jul 18, 2016

$x = {e}^{{e}^{y} - 2}$

#### Explanation:

Calling

$y = {\log}_{e} \left(2 + {\log}_{e} \left(x\right)\right)$

we have also

${e}^{y} = 2 + {\log}_{e} \left(x\right)$

and ${\log}_{e} \left(x {e}^{2}\right) = {e}^{y}$ then

$x {e}^{2} = {e}^{{e}^{y}}$ and finally

$x = {e}^{{e}^{y} - 2}$

So with this procedure we obtained a function $g$ such that

$x = g \left(y\right)$.

Now we can operate

$y = f \left(x\right) = f \left(g \left(y\right)\right) = f \circ g \left(y\right)$ such that

$f \circ g \equiv 1$. Here $g$ is called inverse function regarding $f$

Jul 18, 2016

Inverse function of $f \left(x\right) = y = \ln \left(2 + \ln x\right)$ is
$f \left(x\right) = {e}^{{e}^{x} - 2}$

#### Explanation:

Let $f \left(x\right) = y = \ln \left(2 + \ln x\right)$.

Hence, $2 + \ln x = {e}^{y}$ or

$\ln x = {e}^{y} - 2$ and

$x = {e}^{{e}^{y} - 2}$

Hence inverse function of $f \left(x\right) = \ln \left(2 + \ln x\right)$ is
$f \left(x\right) = {e}^{{e}^{x} - 2}$