Solve for x with exact values (no decimals):

e^(2x) - e^x - 6 = 0

I know how to generally solve this problem - by changing it to quadratic form:
(e^x - 3)(e^x + 2)=0
And seeing that:
e^x = 3
and
e^x = -2 (though this one is not possible)

So i'm left with e^x = 3. Ln both sides:
xln(e) = ln(3)
x = ln(3)

But I can't do it this way since solving ln(3) would require a calculator (and a huge number of decimals no less)

Is there another trick to solving problems such as this for exact values? Thanks