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

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