Undoing ln

Now the log is what is in the way. To solve $\ln\left(x\right) = 2$, read it as a power: the power of e that gives x is 2. So $x = {e}^{2}$, about 7.39.

Raising e to both sides does the same thing by algebra. ${e}^{\ln\left(x\right)}$ is x, because a power of e and ln undo each other. For a log base 3 you would raise 3 instead.

The answer is always positive, as an input to ln must be. Even $\ln\left(x\right) = -1$ has a solution: $x = {e}^{-1}$, about 0.37. A negative log only means x is between 0 and 1.