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

Question

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*}

Answer 1

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}

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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}

Answer 2

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 $\log_a b \log_b a = 1$.

Answer 3

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.