log4x-log4(x+3) = log4(x-2)

Get all logs on the side of the equation where they will be positive

log4x = log4(x+3) + log4(x-2)

The left side is already a single log.  We get the right side
to a single log also.  To do this we remember that the sum of logs
is the log of the product:

log4x = log4[(x+3)(x-2)] 

Now that we have single logs on both sides we can drop the logs

x = (x+3)(x-2)

x = x²+x-6

0 = x²-6

6 = x²

±√6 = x

But since we cannot take logs of negative numbers, we must discard
the negative solution, sinc the original equation has term log4x.
However √6 causes us to only have to take logs of positive solutions, 
so it is a solution.

Therefore the only solution is √6 

Edwin