My teacher says there's no solution to this question, but I'm getting answer and I figured that I probably missed something. Can someone check it?

Question

jojokiw3 · Answer

$$\log_{3} (x-2) - \log_{3} (x+1)= \log_{3} 5 $$

jojokiw3 · Answer

I keep getting $$\frac{ -7 }{ 4 }$$ as the answer for x.

anonymous · Answer

I'm getting the same.. and so does wolfram alpha http://www.wolframalpha.com/input/?i=log3%28x-2%29+-+log3%28x%2B1%29+%3D+log3%285%29 are you sure the question is right?

jojokiw3 · Answer

Yeah I'm sure. I guess the teacher is probably wrong? o_O

jojokiw3 · Answer

I looked at over 20 times lol.

anonymous · Answer

The thing is that wolfram writes '(assuming a complex-valued logarithm)' on the solution for this one, but not for others. let me think why it matters

jojokiw3 · Answer

I don't even know what that is. -_-

anonymous · Answer

Ok, I think I got it.
Think about it this way, is there a real power for 3 that will cause $3^p \le 0$?
No, so n has to be positive in order for $\log_3(n)$  to be able to return a real value.
So if we say our log function has to work only with real numbers and cannot return complex numbers, then we can't take logarithm of non-positive numbers.

So we have:
$$
\log_3(x-2) - \log_3(x+1) = \log_3(5)
$$That means we must have:
$$
\begin{array}{ll|ll}
x-2 > 0 &&& x+1 > 0\
x > 2 &&& x > -1 \
\hline
\end{array}\
\boxed{x>2}
$$But our result is $x = -\frac{7}{4}$ which is not in the valid range for $x$, so there is no solution.

jojokiw3 · Answer

Oh okay. Thanks chum!

anonymous · Answer

You're welcome =)