Help! A logarithm that won't be solved

[Guest]

The problem is -

log(3x+3)-2log(x+1)=2log3

The answer is supposed to be 2/3 and 1 but I keep getting to 1 and 4 - I end up factorising -x2+5x+4

Please help, as I am seriously confused!

 

[Gene]

log(3x+3)-2log(x+1)=2log3
log(3(x+1)/(x+1)^2)=log(3^2)
3/(x+1)=3^2
1/3=x+1
x= - 2/3
That works. I don't see a 1 though. How did you get your equation?

 

[stapel]

Multi-post.

 

[Gene]

One grows confused again. In your other post Eliz said

x = 2/3:
. . . . .log(2 + 3) - 2log(2/3 + 1) ?=? 2log(3)
. . . . .log(5) - 2log(5/3) ?=? log(9)
. . . . .log(5) - log(25/9) ?=? log(9)
. . . . .log((5/1)/(25/9)) ?=? log(9)
. . . . .log(9) ?=? log(9)

But (5/1)/(25/9) = 9/5 not 9
I have to stick with my - 2/3 and disagree with the book's + 2/3.
-------------------
Gene

 

[stapel]

...But (5/1)/(25/9) = 9/5 not 9

D'oh! :roll:

Gene is right, of course. Which reinforces what was said earlier: check the solutions for correctness (and show your work, so others can check, too).

Thanks for catching that, Gene!

Eliz.