Logarithms do not only involve the use of logarithm tables; it also works with the use of the logarithms laws which are interrelated to the laws of indices. Moreover, the laws of indices have equivalent laws of logarithms as we will see in this article.

In this article, we are going to look at the laws of logarithms and examples of all laws with step-by-step solutions. We will likewise drop the worksheet. This will contain quizzes similar to the examples in the article.


The following are the different rules that can be used when solving problems on logarithms:

Product Law

In some clans, this can be called the ‘addition law‘. It is the same, this simply implies that when two logarithms of the same base are added, the result is the product of the two logarithms to their common base. That is,  {\color{Emerald} loga^{x}+ loga^{y}=loga^{xy}}On the other hand, this can be the logarithm of products of two numbers with a common base. Which will result in the addition of the two numbers with the same base. This can be represented by {\color{Emerald} log a^{MN}=log a^{M}+loga^{N}}.

For example, log2^{8}=log2^{(2\times 4)}  which will give log2^{2} + log 2^{4}.

Quotient Law

This is also referred to as the ‘subtraction law‘. It shows that when logarithms are subtracted, the one being subtracted is used to divide the other, all expressed in their common base. That is,

{\color{Emerald} loga^{x}-loga^{y}=loga^{\frac{x}{y}}}

On the other hand, these can be written as {\color{Emerald} loga^{\frac{M}{N}}=loga^{M}-loga^{N}}.

The questions on these two laws can come in any of the mentioned forms. The proper application of the laws is what is required.

For example, log10^{\frac{1000}{100}}=log10^{1000}-log10^{100}.

Power Law

When a logarithm is raised to a certain power, the power is used to multiply the logarithm itself. This law is expressed in the form:  {\color{Emerald} loga^{M^{p}}= Ploga^{M}}

For example, log2^{2^{3}}=3log2^{2}.

Apart from the above laws and more that will be discussed later, there are principles that can be applied when solving questions on the laws of logarithms. We can call them the key logarithm rules. These rules must be known since they will most probably be applied in some of these laws.

Recommended: Standard form of a number


There are two of these rules:

  1. Logarithms to its own Base: The logarithm of any value to its own base is equal to 1. That is, loga^{a}=1. For example, log10^{10}=1, log8^{8}=1, log100^{100}=1 etc.
  2. Logarithm of 1: The logarithm of 1 to any base is equal to zero. Tha is, loga^{1}=0, where a1. For example, log10^{1}.

To prove that log10^{1} is equal to zero, Let log10^{1}=x.

According to a logarithm law, the above becomes 10^{x}=1.

Recall that 10^{0}=1.

So 10^{x}=10^{0}  (the 10 will clear the 10),

∴ x=0

So you see, log10^{1}=0.

We will see the application of these various laws in complex examples, later in this article. Before then, let’s look at other laws of logarithms. Consider the fractional power law.


Fractional Power law

In some cases, a logarithm might have fractions as its powers. These are fractional power logarithms, and it can be solved like the power law. It is represented by,

{\color{Emerald} loga^{M^{\frac{x}{y}}}=\frac{x}{y}loga^{M}}

Which can also be expressed as {\color{Emerald} \frac{xloga^{M}}{y}}

For example, log10^{100^{\frac{3}{2}}}=\frac{3}{2}log10^{100}=\frac{3log10^{100}}{2}

Root Law

This is a special case of fractional power law. In this case, x=1 and the root law is applied. Recall from the laws of indices that, a^{\frac{1}{2}}=\sqrt{a}, a^{\frac{1}{4}}=\sqrt[4]{a} etc. So root law can be expressed as

{\color{Emerald} loga^{M^{\frac{1}{y}}}=loga^{\sqrt[y]{M}}}

Note, this can also be written as  \frac{1}{y}loga^{M}=\frac{loga^{M}}{y}

Whatever method you use in solving questions like this, you will arrive at the same result.

For example, log2^{16^{\frac{1}{4}}}=log2^{\sqrt[4]{16}}.

Reciprocal Law

When the reciprocal of a logarithm is required, the base and the number interchange their positions. That is, {\color{Emerald} loga^{M}=\frac{1}{logM^{a}}}

For example, log2^{8}=\frac{1}{log8^{2}}

Change of Base Law

This law shows that when the base of a logarithm is changed, the initial base is used as a separate logarithm to divide the initial logarithm. All to the new base. That is, {\color{Emerald} loga^{M}=\frac{logx^{M}}{logx^{a}}}

From the above, the initial base divides the initial logarithms which is ‘M‘, all to the same new base ‘x‘.

For example, log100^{1000}=\frac{log10^{1000}}{log10^{100}}.

Finally, let’s consider some examples, explaining the above formulas. As you go through these examples, see how the laws are applied, that will help you assimilate them properly and also apply them when necessary.


We are going to consider a series of examples that will explain the practical application of the above laws.

Example 1: Simplify log8^{16^{-2}}

From the above question, log8^{16^{-2}}= -2log8^{16}  (applying the Power law)

-2log8^{16}=-2(\frac{log2^{16}}{log2^{8}})  (change of base law)

Simplify further by reducing 16 and 8,

-2(\frac{log2^{2^{4}}}{log2^{2^{3}}}) =-2(\frac{4log2^{2}}{3log2^{2}}), Recall one of the rules we discussed above, rule 1, (log2^{2}=1)

-2(\frac{4\times 1}{3\times 1})=-2(\frac{4}{3})=\frac{-2\times 4}{3}


Example 2: Simplify \frac{1}{2}log5^{25}- log5^{0.2}

Firstly, let’s convert the decimal to fractions, 0.2=\frac{2}{10}=\frac{1}{5},


Secondly, lets apply the power law and this law of indices \frac{1}{5}=5^{-1},

\frac{1}{2}\times 2log5^{5}-log5^{5^{-1}}=log5^{5}-(-1)log5^{5}

Applying the rule, log5^{5}=1 and remembering that -\times -=+ gives us


Example 3: Simplify 2log3^{6}+log3^{12}-log3^{16} (WAEC)

Use the product and quotient laws, since all the values have the same base,

log3^{(\frac{36\times 12}{16})}=log3^{\frac{432}{16}}


Applying the Power law,

3log3^{3}  (recall that log3^{3}=1)

3\times 1=3

Example 4: If log2=0.3010, evaluate log 32.

Using the product law,

log10^{32}=log10^{(8\times 4)}

log10^{(8\times 4)}=log10^{8}+log10^{4}

=log10^{2^{3}}+log10^{2^{2}} (Applying the power-law)


since log2=log10^{2}=0.3010



In conclusion, let us learn how to relate logarithms and indicial equations. An indicial equation is an equation involving one or more unknowns as the exponent (or index). The example below will explain this further.

Example 5: Solve the logarithm equation log9^{(\frac{1}{3})^{x}}-log9^{(\frac{1}{9})^{x}}=2

Applying the logarithm power law,



Using the division rule, \frac{1}{3} ÷ \frac{1}{9}=3

So x(log9^{\frac{1}{3}}-log9^{\frac{1}{9}})=2 gives xlog9^{3}=2

Reversing the power law, log9^{3^{x}}=2

Recall that if log2^{x}=6, x=2^{6}. Applying that to the above, we get


3^{x}=3^{2^{2}} (clear the common base, 3)


∴ x=4

Example 6: Solve for x in the logarithm equation log_{3}{x}+log_{3}{(x-8)}=2. Watch the video below for the solution.

For questions on this topic, “Logarithms Laws and Example” download the law of logarithms work Sheet. Click HERE to download.


For insight on this topic and more, get a copy of the textbook “New Track Mathematics for JAMB UTME SERIES 1” It contains comprehensive mathematics step-by-step solutions with JAMB Past Questions and answers.

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September 13, 2021

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