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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
