Proof that $\log_a b \cdot \log_b a = 1$ 
Prove that $\log_a b \cdot \log_b a = 1$

I could be totally off here but feel that I have at least a clue.
My proof is: 
Suppose that $a = b$, then $a^{1} = b$ and $b^{1} = a$ and we are done.
Suppose now that $a \neq b$. We wish to show that: $\log_b a = \frac{1}{\log_a b}$.
\begin{align*}
1 &= \frac{1}{\log_a b} \cdot \frac{1}{\log_b a} \cdot \left(\log_a{b} \cdot \log_b a \right)    \\
   &=  \left(\log_a{b} \cdot \frac{1}{\log_a b} \right) \cdot  \left(\log_b a \cdot \frac{1}{\log_b a}\right)  \\
   &=  \left(\log_b a \cdot \frac{1}{\log_b a}\right) \cdot 1    \\
   &=1 \cdot 1    \\
  &= 1
\end{align*} 
 A: Your argument does not prove the identity. I can substitute any two values for $\log_a b$ and $\log_b a$ and get the "proof"
\begin{align}
1 &= \frac1x \cdot \frac1y \cdot \left( x \cdot y \right) \\
&= \left( x \cdot \frac1x \right)\left( y \cdot \frac1y \right) \\
&= \left( x \cdot \frac1x \right) \cdot 1 \\
&= 1 \cdot 1 \\
&= 1
\end{align}
But we certainly don't have $xy = 1$ for all values of $x$ and $y$. Your manipulations aren't wrong, but you've only shown that $1 = 1$ not that $\log_a b \log_b a = 1$.
A: Method I
Let $\log_ab=x$ and $\log_ba=y$. Then we have
$$a^x=b \quad\text{and}\quad b^y=a$$
So,
\begin{align}
(b^y)^x&=b\\
b^{xy}&=b\\
xy&=1
\end{align}

Method II
Let $\log_ab=x$. Then we have
\begin{align}
a^x&=b\\
\log_b(a^x)&=\log_bb\\
x\log_ba&=1
\end{align}
A: That could be made into something that works but I think that route is a little more than necessary.
When trying to establish an equality, in this case $\log_ab \cdot \log_ba = 1$, it's good form not to mix the two sides together.  In other words, don't divide both sides by $\log_ab$.
This can be directly and quickly shown by using the change of base formula.
$\log_ab = \dfrac{\ln b}{\ln a}$ and $\log_ba = \dfrac{\ln a}{\ln b}$.  Multiply them together and...
Side note:  Care must be taken that we don't divide by zero anywhere.  Basically this means (in your method and my method) that $a \ne 1$ and $b \ne 1$.  But we should have this anyway from the definition of the base of a logarithm.
