In the **Product of Logarithm of A and Logarithm of B**, we have seen that $latex \log AB = \log A + \log B$. In this post, we will see prove the quotient law of logarithm. We will show that

$latex \displaystyle\frac{\log A}{\log B} = \log A – \log B$.

In proving, we will use the connection between the logarithm notation and exponent notation. Recall in the previous post that $latex \log_{10}x = c$ if and only if $latex 10^c = x$.

**Theorem**

$latex \displaystyle\frac{\log A}{\log B} = \log A – \log B$.

**Proof**

Let $latex \log_{10} A = x$ and $latex \log_{10}B = y$.

In exponent notation, these are equivalent to $latex A = 10^x$ and $latex B = 10^y$.

Now, $latex \displaystyle\frac{A}{B} = \frac{10^x}{10^y} = 10^{x-y} $.

Getting the logarithm of both sides, we have

$latex \log \displaystyle\frac{A}{B} = \log 10^{x – y}$

which means that

$latex \log \displaystyle \frac{A}{B} = x – y$.

But $latex x = \log_{10}A$ and $latex y = log_{10} B$.

Therefore, $latex \log \frac{A}{B} = \log A – \log B$. $latex \blacksquare$.

This is a good explanation of the quotient law for logarithms. I have been meaning to post a series about logarithms on my site as well.