Question 895516
<pre>
log<sub>4</sub>x-log<sub>4</sub>(x+3) = log<sub>4</sub>(x-2)

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

log<sub>4</sub>x = log<sub>4</sub>(x+3) + log<sub>4</sub>(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:

log<sub>4</sub>x = log<sub>4</sub>[(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

±&#8730;<span style="text-decoration: overline">6</span> = x

But since we cannot take logs of negative numbers, we must discard
the negative solution, sinc the original equation has term log<sub>4</sub>x.
However &#8730;<span style="text-decoration: overline">6</span> causes us to only have to take logs of positive solutions, 
so it is a solution.

Therefore the only solution is &#8730;<span style="text-decoration: overline">6</span> 

Edwin</pre>