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September 16, 2014

September 16, 2014

Posted by **Samir** on Thursday, January 24, 2008 at 11:29am.

Both of them are base 10

- advanced fuctions -
**jAKE1214**, Thursday, January 24, 2008 at 1:22pmI haven't done log for a while, but can't you just use subsitution and set up an equation like this?

log(x-2)=1-log(x+1) and then solve for x?

log(x-2)+log(x+1)=1 and I think there's a rule that let you combine those... like multiplying them or something and go from there. Or if you have a graphing calculator, just use that to find the intersections for checking your answer.

- advanced fuctions -
**Reiny**, Thursday, January 24, 2008 at 3:35pmJake is on the right track, from his

log(x-2)+log(x+1)=1

log((x-2)(x+1))=1

log(x^2 - x - 2) = 1

x^2 - x - 2 = 10^1

x^2 - x - 12 = 0

(x-4)(x+3) = 0

x = 4 or x = -3, but x > -1 for log (x+1) to be defined,so

x = 4

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