Undoing ln

Now the log is what is in the way. To solve ln(x) = 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(x)) 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(x) = -1 has a solution: x = e^(-1), about 0.37. A negative log only means x is between 0 and 1.