Thread: Log equation question...

Hi all. I'm having difficulty seeing how my solution to a problem is wrong. Here it is, along with my attempt:

log(2x+1) = 1-log(x-1)
log(2x+1) + log(x-1) = 1
log((2x+1)(x-1)) = 1
log(2x^2-x-1) = 1
2x^2-x-1 = 10
(2x+1)(x-1)=10
2x=9, x=9/2
x=11

I'm sure it's a case of "oh, duh." :P Thanks in advance...

Originally Posted by NJDesigner

Hi all. I'm having difficulty seeing how my solution to a problem is wrong. Here it is, along with my attempt:

log(2x+1) = 1-log(x-1) <-- don't forget: x > 1
log(2x+1) + log(x-1) = 1
log((2x+1)(x-1)) = 1
log(2x^2-x-1) = 1
2x^2-x-1 = 10
...
I'm sure it's a case of "oh, duh." :P Thanks in advance...

This is a quadratic equation. To solve it use the quadratic formula:



You'll get 2 solutions but only one satisfies the original equation.

Thanks! I had a feeling actually but I wasn't confident about it. I'm looking how to give points now (newbie here)