$\log_{b}{\Big(\dfrac{m}{n}\Big)}$ $\,=\,$ $\log_{b}{m}-\log_{b}{n}$

The quotient rule is another most useful logarithmic identity, which states that logarithm of quotient of two quotients is equal to difference of their logs. The property of quotient rule can be derived in algebraic form on the basis of relation between exponents and logarithms, and quotient rule of exponents.

$m$ and $n$ are two quantities, and express both quantities in product form on the basis of another quantity $b$.

The total multiplying factors of $b$ is $x$ and the product of them is equal to $m$.

$m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$

$\implies m \,=\, b^{\displaystyle x}$

In the same way, the total multiplying factors of $b$ is $y$ and the product of them is equal to $n$.

$n$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle y \, factors}$

$\implies n \,=\, b^{\displaystyle y}$

Thus, the two quantities are written in exponential notation as follows.

$(1) \,\,\,\,\,\,$ $m \,=\, b^{\displaystyle x}$

$(2) \,\,\,\,\,\,$ $n \,=\, b^{\displaystyle y}$

On the basis of mathematical relation between exponents and logarithms, the quantities in exponential form can be written in logarithmic form as follows.

$(1) \,\,\,\,\,\,$ $b^{\displaystyle x} \,=\, m$ $\,\, \Leftrightarrow \,\,$ $\log_{b}{m} = x$

$(2) \,\,\,\,\,\,$ $b^{\displaystyle y} \,=\, n$ $\,\,\,\, \Leftrightarrow \,\,$ $\log_{b}{n} = y$

Divide the quantity $m$ by $n$ to get the quotient of them mathematically.

$\dfrac{m}{n}$

Actually, the values of the quantities $m$ and $n$ in exponential notation are $b^{\displaystyle x}$ and $b^{\displaystyle y}$ respectively.

$\implies \dfrac{m}{n} \,=\, \dfrac{b^{\displaystyle x}}{b^{\displaystyle y}}$

According to the quotient rule of exponents, the quotient of exponential terms whose base is same, is equal to the base is raised to the power of difference of exponents.

$\implies \dfrac{m}{n} \,=\, b^{\,({\displaystyle x}\,-\,{\displaystyle y})}$

Take $d = x-y$ and $q = \dfrac{m}{n}$. Then, write the equation in terms of $d$ and $q$.

$\implies q = b^{\displaystyle d}$

Now write the equation in log form.

$\implies \log_{b}{q} = d$

Replace the original values of the quantities $d$ and $q$.

$\implies \log_{b}{\Big(\dfrac{m}{n}\Big)} = x-y$

The logarithm of quotient of two quantities $m$ and $n$ to the base $b$ is equal to difference of the quantities $x$ and $y$.

In fact, $x \,=\, \log_{b}{m}$ and $y \,=\, \log_{b}{n}$. So, replace them to obtain the property for the quotient rule of logarithms.

$\,\,\, \therefore \,\,\,\,\,\, \log_{b}{\Big(\dfrac{m}{n}\Big)}$ $\,=\,$ $\log_{b}{m}-\log_{b}{n}$

It has proved that the logarithm of quotient of two quantities to a base is equal to difference their logs to the same base. The fundamental law is also called as division rule of logarithms and used as a formula in mathematics.

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