What is the nth root of a number

Question

Given $$x = y^\frac{1}{ln y}$$ $$x = ?$$

Is x is the approximation of e ?

Answer 1

Please note that $ \ln(1) = 0 $ and $ \ln(0)$ is undefined thus,

For $ \{ x | x \ne 0, x\ne 1 , x \in \mathbb{C} \} $,
$ x $ is exactly equal to $e$

Proof: $$ \because \ln(x) $$ is the inverse of $ e^x $ , $$\ln(e^x) = x $$ and $$ e^{\ln(x)} = x $$ $$ \therefore y = e^{ln(y)} $$

$$ y^{\frac{1}{\ln(y)}} = (e^{ln(y)}) ^ {\frac{1}{\ln(y)}} $$ $$ = e^{\ln(y) * \frac{1}{\ln(y)}} = e^{1} $$

Thus, $ x = y^{\frac{1}{\ln(y)}} = e $