# How do you solve lnx=2?

$x = 7.3891$
If ${a}^{n} = b$, we have ${\log}_{a} b = n$, where $a$ is called base. When base is $a = e$, we have Napier's logarithm and we need not write base and we just write it as $\ln b = n$ which is equivalent to $b = {e}^{n}$.
Hence, $\ln x = 2$, can be written as $x = {e}^{2}$ and as ${e}^{2} = 7.3891$, $x = 7.3891$