Hi All!

I've been having trouble with this one logarithmic problem.

It starts like this...

Solve for x

logx + log (x+1) = log12

I have

log (x)(x+1) =12
log x^2 + 1x =12

x^2 + 1x =12
x^2 + 1x -12 =0

(x )(x )

I can't figure it out from here. I'm stumped! :confused:

Thank you for all your help!

What are the factors of 12?

They are 1, 2, 3, 4, 6, 12.

Now which of these will multiply to give 12?

1 and 12, 2 and 6, 3 and 4.

Which ones will fit the equation? That, you'll have to look fot it. But I'm sure that you already know the answer. If not, post.

Apart from that, everything is OK. Good job!:)

after the equation:
x^2+x-12=0
proceed as:
x^2+4x-3x-12=0
x(x+4)-3(x+4)=0
(x-3)(x+4)=0
so you get two values of x i.e. +3 and -4

now if you put -3 in the given equation you get:
log(-4)+log(-4+1)=log12
but here is one important point i.e. log cannot have negative values in it so,
x=-4 is wrong value of x,
hence x=3 is the answer...

Hum, did you know that you are not really supposed to give direct answers here. Guiding is more advisable, except in cases where this cannot be applied. This question can be explained but easy if you practise a lot of similar questions. Next time, try to guide, instead of directly answering similar questions, OK?