How do I sole
ln(x^(2) - x - 2) = 2 + ln(x+1)
This is as far as I have gotten
ln((x^(2) - x - 2)/(x+1)) = 2
(x^(2) - x - 2)/(x+1) = e^(2)
then I divided because I was too lazy to do algebra
(x+1)(x-2) = e^(2)
=
x^(2)-x-2-e^(2) = 0
I'm not allowed to use a calculator
I was thinking I could just sub these numbers into the quadratic formula to come up with solutions would that be the right course of action?

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\[x^2-x-2=(x+1)(x-2)\] so you can cancel either first or last

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