Proving that $\log_b(r^s) = s\log_b(r)$ The Question
If $b,r,s \in \mathbb{R^+}$ prove that $\log_b(r^s) = s\log_b(r)$
My Work
1) $\log_b(r^s)$ 
2) $s$ can be expressed as the sum of an integer part $n$ and a real part $m$: $s = m + n$
3) $\log_b(r^{n+m})$
4) $\log_b(r^nr^m)$
5) $\log_b(r^n) + \log_b(r^m)$
6) $\log_b(r\cdot r\cdot r \cdots r) + \log_b(r^m)$
7) $\log_b(r) + \log_b(r) + \log_b(r) + \cdots + \log_b(r) + \log_b(r^m)$
8) $n\log_b(r) + \log_b(r^m)$
Where I am Having Trouble
I'm having trouble getting the $m$ in front of the second summand which I feel is necessary for the theorem. How can I finish off this proof? I was given this rule as a secondary school student with no proof and would like to have it explained. 

$$\log_b(x) = y \iff b^y=x$$
Using the definition we note that
$$b^{s\log_b(r)}=(b^{\log_b(r)})^s=r^s$$
So $\log_b(r^s) = s\log_b(r)$.
