I need to prove the logarithmic power rule: $$\log_b(x^r) = r \cdot \log_b(x) \hspace{2.3cm}$$
I have seen a large number of sources citing a similar proof, which goes like this:
\begin{align*} \text{let} \ \ m &= \log_b(x), \\ x &= b^m \\ x^r &= (b^m)^r \\ \log_b(x^r) &= \log_b((b^m)^r) \\ &= \log_b(b^{mr})& \text{(5)}\\ &= mr & \text{(6)}\\ &= rm \\ \log_b(x^r) &= r \cdot \log_b(x) & \text{Q.E.D} \\ \end{align*}
The problem I have with this proof is that in going from steps 5 to 6, I implicitly justify them as $$\hspace{0.7cm} \log_b(b^{mr}) = mr \cdot \log_b(b) = mr \cdot 1 = mr $$
In this case, it appears that the conclusion to be proved has been used, and the proof is invalid because it is simply a case of circular reasoning.
Is everybody wrong, or have I missed something?
EDIT: If the proof is in fact broken, can someone please suggest how it could be fixed?
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=y\iff x=\log_{a}y$ ($=\log_{a}a^x$)
It is a bit like saying $\log_{10}{1000}=3$ and that you don't need any fancy properties; it follows from the definition.