Use the Laws of Logarithms to combine the expression.

Hmm.  I'm not sure Maple agrees with me.  I based my reply above on reasoning about the two terms separately, and in this Maple agrees with me - see the first plot below (I've just used the natural log; a different base only alters the scaling, not the domain).

However, when I get Maple to plot the difference of the two terms it gives the following (which is identical to ln(x+8)):

So, Chris, it looks like the mathematical software big-hitters (Mathematica and Maple) think it's ok to combine terms before worrying about domain validity!

$$log_8 \frac{(x-8)(x+8)}{x-8}=log_8 (x+8) \qquad \mbox{where }x>-8\;\; and \;\;x\ne 8$$

I may be incorrect about this, Melody, but I believe that x must be greater than 8. The answer has to make sense in the original problem, as well. If x is only greater than -8, the second expression in the original problem is undefined for (-8,8].

No mine is correct Chris

http://www.wolframalpha.com/input/?i=log%28base8%29%28%28x%5E2-64%29%2F%28x-8%29%29

x cannot equal 8 because you cannot divide by 0

x+8>0 because you cannot find the log of a number that is less than or equal to 0

so

x>-8 and x not equal to +8

If we are keeping to the Real numbers only, then log(x-8) isn't defined unless x is strictly greater than 8, so Chris is correct in this situation.

However, if we allow Complex numbers then we have log(x^2-64) - log(x-8) = log(x+8) + log(x-8) - log(x-8) = log(x+8) for all x except x = -8 and x = 8.

Yes I appologise, I can see that now that you have pointed it out.  

So, thank you Alan and my appologies to you Chris.