Question

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?

Answer 1

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

Answer 2

The script displaying Latex type is crashing blah

Answer 3

probably easier to do it first,
\[\ln((x+1)(x-2))=\ln(x+1)+\ln(x+2)=2+\ln(x+2)+\]
\[\ln(x-1)=2\]
\[x-2=e^2\] etc

Answer 4

if you can't see the latex i say it in english
the x + 1 cancels, you don't get a quadratic

Answer 5

I will brb going to restart my browser so I can read what you wrote