Logarithm Quotient Law


$\log_{\displaystyle b} \Bigg( \dfrac{m}{n} \Bigg) = \log_{\displaystyle b} m \,–\, \log_{\displaystyle b} n$

It is read as the logarithm of division of $m$ by $n$ to base $b$ is equal to logarithm of $m$ to base $b$ minus logarithm of $n$ to base $b$. It is useful to find the value of logarithm of division of any two numbers by the subtraction of logarithms of them.


$m$ is a number and it is expressed as multiplying factors of another number $b$. The total multiplying factors of $b$ is $x$. The relation between three of them can be expressed in exponential notation.

$m = b^{\displaystyle x}$

Express the same relation in logarithmic form.

$\log_{\displaystyle b} m = x$

Similarly, $n$ is another number and it is also expressed as multiplying factors of same number $b$. In this case, the total multiplying factors of $b$ is $y$ to obtain the $n$.

$n = b^{\displaystyle y}$

Express the same exponential relation in logarithmic function form.

$\log_{\displaystyle b} n = y$

Take ratio of $m$ to $n$.

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

Use division Rule of exponential terms having same base to obtain the quotient of them.

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

Take, the quotient of ratio $m$ to $n$ is $q$, therefore $q = \dfrac{m}{n}$ and also take $x-y = d$

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

Express it in logarithmic system.

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

Now replace, the values of $q$ and $d$ by their actual values.

$log_{\displaystyle b} \dfrac{m}{n} = x -y$

Replace the actual values of $x$ and $y$ in terms of logarithms to obtain the quotient property of the logarithms in mathematical form.

$\therefore \,\,\,\,\, \log_{\displaystyle b} \Bigg(\dfrac{m}{n}\Bigg) = \log_{\displaystyle b} m \,–\, \log_{\displaystyle b} n$


$\log_{\displaystyle 10} \Bigg( \dfrac{3}{2} \Bigg)$

$\implies \log_{\displaystyle 10} \Bigg( \dfrac{3}{2} \Bigg) = \log_{\displaystyle 10} 1.5$

$\implies \log_{\displaystyle 10} \Bigg( \dfrac{3}{2} \Bigg) = 0.176091259 \cdots \approx 0.1761$

The value of $\log_{\displaystyle 10} \Bigg( \dfrac{3}{2} \Bigg)$ is equal to $0.1761$

$\log_{\displaystyle 10} 3 = 0.477121254 \cdots \approx 0.4771$

$\log_{\displaystyle 10} 2 = 0.301029995 \cdots \approx 0.301$

Subtract $\log_{\displaystyle 10} 2$ from $\log_{\displaystyle 10} 3$

$\log_{\displaystyle 10} 3 \,-\, \log_{\displaystyle 10} 2 = 0.4771 – 0.301$

$\implies \log_{\displaystyle 10} \,-\, \log_{\displaystyle 10} 2 = 0.1761$

Now compare the values to verify this identity.

$\therefore \,\,\,\,\, \log_{\displaystyle 10} \Bigg( \dfrac{3}{2} \Bigg)$ $=$ $\log_{\displaystyle 10} 3 \,-\, \log_{\displaystyle 10} 2$ $=$ $0.1761$

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