solve for x. logx+log(x-1)=log(4x)

Question

aum · Answer

$\large \log(A) + \log(B) = \log(AB)$
Use the above to combine the two logs on the left hand side.
Then use the fact that if log(C) = log(D), then C = D

samsan9 · Answer

log(x) + log(x - 1) = log(4x)
log[x(x - 1)] = log(4x)
x(x - 1) = 4x
x^2 - x - 4x = 0
x^2 - 5x = 0
x(x - 5) = 0
x = 0 or 5  is this right?

aum · Answer

Correct.  But you have to see if there are any extraneous solutions.
If we put x = 0 in the original equation we realize it is no good because log is defined only for numbers greater than 0.  zero and negative numbers are not allowed for log.
So x = 0 is not a valid solution.

But x = 5 is a valid solution.

samsan9 · Answer

oh ok so no matter what anything zero or less is  wrong ?

aum · Answer

log(x) is defined only for x > 0.

aum · Answer

So if you get x = 0 or x = -3 or x = -100 you have to throw it away because log(x) does not exist for those x.