# Proof of Logarithmic Product identity

The logarithm of the product of two or more quantities is equal to the sum of their logarithms as per the product rule of the logarithms. The product property of the quantities in logarithmic form is written mathematically as follows.

$\log_{b}{(m \times n)}$ $\,=\,$ $\log_{b}{m}+\log_{b}{n}$

It is time to learn how to derive the product law of the logarithms in algebraic form.

### Write the quantities in Exponential form

Let $m$ and $n$ express two quantities in algebraic form, and they both are factored on the basis of a literal quantity $b$. Therefore, the literal quantities $m$ and $n$ can be written in terms of the literal quantity $b$.

$(1).\,\,$ $m$ $\,=\,$ $b \times b \times b \times \cdots \times b$

$(2).\,\,$ $n$ $\,=\,$ $b \times b \times b \times \cdots \times b$

Let’s assume that the number of factors in $b$ to express the quantity $m$ in the product form is denoted by a variable $x$, and the number of factors in $b$ to express the quantity $n$ in the product form is denoted by a variable $y$.

$(1).\,\,$ $m$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle x \, factors}$

$(2).\,\,$ $n$ $\,=\,$ $\underbrace{b \times b \times b \times \ldots \times b}_{\displaystyle y \, factors}$

Use the exponentiation to write the quantities in product form in exponential notation.

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

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

### Express the quantities in Logarithmic form

Now, use the mathematical relation between the logarithms and exponents to convert the exponential equations into the logarithmic equations.

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

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

Therefore, the values of $x$ and $y$ in logarithmic form can be written as follows.

$(1).\,\,$ $x \,=\, \log_{b}{m}$

$(2).\,\,$ $y \,=\, \log_{b}{n}$

### Find the Product of quantities by multiplication

Now, let us multiply the literal quantities $m$ and $n$ to get the product of them mathematically.

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

Look at the exponential functions on the right hand side of the equation. They both have the same base. So, the product of them can be obtained as per the same base product rule of the exponents.

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

The product of quantities $m$ and $n$ can be written in dot product form or simply $mn$ in mathematics.

$\,\,\,\therefore\,\,\,\,\,\,$ $m.n \,=\, b^{\,{\displaystyle x}\,+\,{\displaystyle y}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $mn \,=\, b^{\,{\displaystyle x}\,+\,{\displaystyle y}}$

### Convert the exponential equation into Logarithm

Let’s denote $s \,=\, x+y$ and $p \,=\, m.n$ for our convenience. Now, write the exponential equation in terms of $s$ and $p$.

$\implies$ $p \,=\, b^{\displaystyle s}$

The above exponential equation can be now transformed into a logarithmic equation by using the mathematical relationship between the exponents and logarithms.

$\implies$ $\log_{b}{p} \,=\, s$

Now, replace the literal quantities $s$ and $p$ in the above logarithmic equation by their actual values.

$\implies$ $\log_{b}{(m.n)} \,=\, x+y$

We have derived that the value of $x$ is logarithm of $m$ to the base $b$ and the value of $y$ is logarithm of $n$ to the base $b$.

$(1).\,\,$ $x\,=\,\log_{b}{(m)}$

$(2).\,\,$ $y\,=\,\log_{b}{(n)}$

Now, substitute the $x$ and $y$ by their corresponding values in the logarithmic equation.

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{b}{(m.n)}$ $\,=\,$ $\log_{b}{(m)}+\log_{b}{(n)}$

We have successfully proved that the logarithm of the product of $m$ and $n$ to the base $b$ is equal to the sum of the logarithm of $m$ to base $b$ and logarithm of $n$ to base $b$.

It is called the product logarithmic identity and it can be extended to more than two factors too.

$\,\,\,\therefore\,\,\,\,\,\,$ $\log_{b}{(m.n.o.p\cdots)}$ $\,=\,$ $\log_{b}{m}$ $+$ $\log_{b}{n}$ $+$ $\log_{b}{o}$ $+$ $\log_{b}{p}$ $+$ $\cdots$

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Jun 26, 2023

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