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.

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