Is this proof with logarithmic exponentials correct?

Question

I was unsure of this proof and some of the log rules I applied, could you check my proof and tell me if this proof is correct and if not, then what specifically is incorrect about the proof?

$\implies \bigl( \log_c b \bigr) \log a = \bigl( \log_c a \bigr) \log b$

$\implies \Bigl( \frac{\log b}{\log c} \Bigr) \log a = \Bigl( \frac{\log a}{\log c} \Bigr) \log b$

$\log c$, on both sides, cancel out (?):

$\implies (\log b) \log a = (\log a) \log b$

$\implies \bigl( \log b - \log a \bigr) \log a = \log b$

$\implies \Bigl( \frac{\log b}{\log a} \Bigr) \log a = \log b$

$\implies \dfrac{\log b}{\log a} = \dfrac{\log b}{\log a}$

Hence the left hand side $=$ right hand side of the equation. I realise I haven't used proof writing language or stated the rules I have applied, I am just trying to wrap my head around how to show this proof first. Am I on the right track here?

Answer 1

You are going through the right motions, more-or-less. I have three comments.

1. The step where you subtract logarithms neither follows from the line above or the line below. It seems like a "mutation" in a an otherwise logically connected sequence of equations.
2. You are dividing by various quantities that could be zero ($\log 1 = 0$, for example). You have to justify how to deal with special cases if they arise. I reread the hypotheses in the statement. With $a, b, c > 1$, there are no division by zero issues.
3. As mentioned in the comments, the argument essentially flows in the wrong direction. Fortunately, for each of your implications, the converse statement is also true. So, to polish the proof, you would write those equations in the reverse order. Begin with something obviously true, and deduce finally the statement quoted at the top.

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Here is what it looks like (modulo explaining how each equation follows from the previous).

$$ \begin{align} \log b \log a &= \log a \log b \\ \frac{\log b}{\log c} \log a &= \frac{\log a}{\log c} \log b \\ \log_c b \cdot \log a &= \log_c a \cdot \log b \\ \log \bigl( a^{\log_c b} \bigr) &= \log \bigl( b^{\log_c a} \bigr) \\ a^{\log_c b} &= b^{\log_c a} \end{align} $$