I 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.
Determine the intersection point of the curves y=log(x-2) and y=1-log(x+1).
Both of them are base 10
2 answers
Jake 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
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