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*}

  • $\begingroup$ This is just change of base rule. $\endgroup$ Jun 15, 2017 at 13:54
  • 4
    $\begingroup$ You starts with $1$ and ends with $1$. So, you have $1=1$. How does this related to the identity? $\endgroup$
    – CY Aries
    Jun 15, 2017 at 13:54
  • 4
    $\begingroup$ Can't follow your calculation. But the point is a simple one: $\log_ab\times \log_ba = \log_b\,a^{\log_a b}=\log_b b =1 $ $\endgroup$
    – lulu
    Jun 15, 2017 at 13:55
  • $\begingroup$ How can you assume a=b? This is only one case and not a proof. $\endgroup$
    – Suprabha
    Jun 23, 2017 at 8:42

3 Answers 3


Method I

Let $\log_ab=x$ and $\log_ba=y$. Then we have

$$a^x=b \quad\text{and}\quad b^y=a$$


\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}


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 $\log_a b \log_b a = 1$.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.