Math Doubts

Product Rule of Logarithms

Formula

$\log_{b}{(xy)}$ $\,=\,$ $\log_{b}{x}$ $+$ $\log_{b}{y}$

What is the product rule of logarithms?

The logarithm of a product is equal to the sum of the logarithms of each factor.

This is the definition of the product law of logarithms, one of the fundamental laws of logarithms. It is used as a formula in mathematics to convert multiplication into addition and is widely used in simplifying logarithms.

In arithmetic, the product of two or more numbers is found by multiplying them together. However, the log of a product cannot be evaluated in the same way.

Example

In arithmetic, $2 \times 3 \,=\, 6$

But, $\log{(2 \times 3)}$ $\,\ne\,$ $\log{(2)}$ $\times$ $\log{(3)}$

To evaluate the logarithm of product, we use a special logarithmic identity called the product law of logarithms.

Examples of Product rule of logarithms

Now, let’s understand the logarithm product rule through a simple and clear example.

Example

$\log_{2}{(8)}$ $\,=\,$ $3$

As per the definition of the logarithm, the logarithm of $8$ to base $2$ is $3$. The number $8$ can be expressed as the product of its factors $2$ and $4$.

$\implies$ $\log{(2 \times 4)}$ $\,=\,$ $3$

Now, let’s find the logarithms of the numbers $2$ and $4$ to base $2$.

  1. $\log_{2}{(2)}$ $\,=\,$ $1$
  2. $\log_{2}{(4)}$ $\,=\,$ $2$

Finally, add the logarithms of both numbers to determine the sum of their logs mathematically.

$\implies$ $\log_{2}{(2)}$ $+$ $\log_{2}{(4)}$ $\,=\,$ $1$ $+$ $2$

$\,\,\,\,\therefore\,\,\,\,$ $\log_{2}{(2)}$ $+$ $\log_{2}{(4)}$ $\,=\,$ $3$

Now, compare both results to understand how the product rule for logarithms works.

$\,\,\,\,\therefore\,\,\,\,$ $\log_{2}{(2 \times 4)}$ $\,=\,$ $\log_{2}{(2)}$ $+$ $\log_{2}{(4)}$ $\,=\,$ $3$

The above numerical example explains why the log of product is equal to the sum of the logarithms of its factors. This logarithmic property is called the product rule in logarithms and the following are some more examples for your better understanding.

  1. $\log_{3}{(10)}$ $\,=\,$ $\log_{3}{(2)}$ $+$ $\log_{3}{(5)}$
  2. $\log_{5}{(42)}$ $\,=\,$ $\log_{5}{(2)}$ $+$ $\log_{5}{(3)}$ $+$ $\log_{5}{(7)}$
  3. $\log_{4}{(3a)}$ $\,=\,$ $\log_{4}{(3)}$ $+$ $\log_{4}{(a)}$

How to Write the Log Product Rule

The product property of logarithms is expressed mathematically as follows.

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

The product of $x$ and $y$ is written as $xy$ or $x.y$ in mathematics.

Domain Conditions

It is one of the most commonly used logarithmic identities in mathematics, but it is important to understand the conditions under which the logarithmic product rule applies.

  1. $b$ denotes a quantity ($b > 0$ and $b \ne 1$) and the base of logarithm.
  2. $x$ and $y$ represent two quantities ($x > 0$ and $y > 0$).
  3. The product rule is valid only when all arguments of the logarithm are positive.

This explains how the log of a product expands into the sum of the logs of its factors.

Extension of the Log Product Rule

The product law for logarithms is not limited to two quantities; it can also be applied to find the logarithm of a product of more than two quantities.

$\log_{b}{(xyz\cdots)}$ $\,=\,$ $\log_{b}{(x)}$ $+$ $\log_{b}{(y)}$ $+$ $\log_{b}{(z)}$ $+$ $\cdots$

The explanation above helps you to understand what the logarithm product identity is, and now it’s time to explore this identity in more depth.

Derivation of the Log Product Law

Let’s learn how to prove that the logarithm of a product is equal to the sum of the logarithms of its factors.

You have learned what the product law in logarithms is, studied its proof, and practiced solving examples to understand how this logarithmic formula works in mathematics.

Properties of Logarithms

Learn similar logarithmic properties that make it easier to simplify expressions and solve problems.

$\log_{b}{\Big(\dfrac{x}{y}\Big)}$ $\,=\,$ $\log_{b}{(x)}$ $-$ $\log_{b}{(y)}$

$\log_{b}{\big(x^n\big)}$ $\,=\,$ $n \times \log_{b}{(x)}$

$\log_{b}{(x)}$ $\,=\,$ $\dfrac{\log_{c}{(x)}}{\log_{c}{(b)}}$

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