# Math Help - Solve for x logarithm

1. ## Solve for x logarithm

Attached is the question and possible solution. Is it correct?

2. $log(x+99)-log(3x-2)=2$

Rewrite:

$log\left(\frac{x+99}{3x-2}\right)=2$

$\frac{x+99}{3x-2}=100$

$x+99=300x-200$

$-299x=-299$

$x=1$

3. It is not correct. Should be :

log(x+99) - log(3x-2) = 2
log( x+99 / 3x-2 ) = 2
x+99 / 3x-2 = 10^2
x+99 = 100(3x-2)
x+99 = 300x - 200
299x = 299
x = 1