# The Quotient of Logarithm of A and Logarithm of B

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

$\displaystyle\frac{\log A}{\log B} = \log A - \log B$.

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

Theorem

$\displaystyle\frac{\log A}{\log B} = \log A - \log B$.

Proof

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

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

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

Getting the logarithm of both sides, we have

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

which means that

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

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

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

## One thought on “The Quotient of Logarithm of A and Logarithm of B”

1. 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.