Question

How do you solve `lnx=2`?

Answer 1

`x=7.3891`

Explanation:

If `a^n=b`, we have `log_ab=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 `lnb=n` which is equivalent to `b=e^n`.

Hence, `lnx=2`, can be written as `x=e^2` and as `e^2=7.3891`, `x=7.3891`