anonymous
  • anonymous
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?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[x^2-x-2=(x+1)(x-2)\] so you can cancel either first or last
anonymous
  • anonymous
The script displaying Latex type is crashing blah
anonymous
  • anonymous
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

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anonymous
  • anonymous
if you can't see the latex i say it in english the x + 1 cancels, you don't get a quadratic
anonymous
  • anonymous
I will brb going to restart my browser so I can read what you wrote
anonymous
  • anonymous
|dw:1326938853539:dw|
anonymous
  • anonymous
wow I can't believe I didn't see that lol, I need to get some sleep thanks :D
anonymous
  • anonymous
that is the ugliest looking 2 i have ever seen!
anonymous
  • anonymous
lol
anonymous
  • anonymous
next time I will factor first
anonymous
  • anonymous
@pross, i think you turned to 2 into an 8
anonymous
  • anonymous
heh, my mouse didn't let go....
anonymous
  • anonymous
yeah i made a typo too, and put \[x-1\] when it should have been \[x-2\] but the method is good
anonymous
  • anonymous
ah now i see what you are talking about...first 2 looked like and 8 and i kept it in the next line.....oops...
anonymous
  • anonymous
you are right...should be x=2+e^2

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