## anonymous 4 years ago 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?

1. anonymous

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

2. anonymous

The script displaying Latex type is crashing blah

3. 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

4. anonymous

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

5. anonymous

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

6. anonymous

|dw:1326938853539:dw|

7. anonymous

wow I can't believe I didn't see that lol, I need to get some sleep thanks :D

8. anonymous

that is the ugliest looking 2 i have ever seen!

9. anonymous

lol

10. anonymous

next time I will factor first

11. anonymous

@pross, i think you turned to 2 into an 8

12. anonymous

heh, my mouse didn't let go....

13. anonymous

yeah i made a typo too, and put $x-1$ when it should have been $x-2$ but the method is good

14. 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...

15. anonymous

you are right...should be x=2+e^2