# Indices

Last updated: September 21, 2021

## How can you manipulate indices?

The index is the power or exponent that is raised to a number or a variable. For example, in number $$2^{4}$$, $$4$$ is the index of $$2$$. The plural form of index is indices. In algebra, we come across constants and variables. The constant is a fixed value, whereas a variable quantity whose values vary or its value can be changed. In this article, we will discuss the laws/rules of the indices along with formulas and solved examples.

## E1.7A: Understand the meaning of indices (fractional, negative and zero) and use the rules of indices.

A number or a variable may have an index.  The index of a variable (or a constant) is a value raised to the power of the variable. The indices are also known as powers or exponents. It shows the number of times a number has to be multiplied.  It is represented in the form:$$a^{m}=a\times a\times a\times a \cdots \times a$$ (m times), where $$a$$ is the base, and $$m$$ is the index.

The index says that a particular number (or base) is multiplied by itself, the number of times equal to the index raised to it.

For example:  $$2^{3}=2\times 2\times 2=8$$

Law of Indices

There are some fundamental rules or laws of indices necessary to understand before we start with indices. These laws are used while performing algebraic operations on indices and solving the algebraic expressions.

Rule 1: If a constant or variable has an index of $$0$$, the result will be equal to one, regardless of any base value.

$$a^{0}=1$$

For example: $$5^{0}=1$$

Rule 2: If the index is a negative value, it can be shown as the reciprocal of the positive index raised to the same variable.

$$a^{-m}=\frac{1}{a^{m}}$$

For example: $$8^{-3}=\frac{1}{8^{3}}$$

Rule 3: To multiply two variables with the base, we need to add its powers and raise them to that base.

$$a^{m}\times a^{n}=a^{m+n}$$

For example

$$2^{3}\times 2^{2}=2^{3+2}$$

$$2^{5}=32$$

Rule 4: To divide two variables with the same base, we need to subtract the power of the denominator from the power of the numerator and raise it to that base.

$$\frac{a^{m}}{a^{n}}=a^{m-n}$$

For example:

$$\frac{10^{4}}{10^{2}}=10^{4-2}$$

$$10^{2}=100$$

Rule 5: When a variable with the same index is raised with a different index, both the indices are multiplied together raised to the power of the same.

$$\left(a^{m} \right )^{n}=a^{mn}$$

For example: $$\left(2^{1} \right )^{2}=2^{2}=4$$

Rule 6: When two variables with different bases but the same indices are multiplied, we have to multiply its base and raise the same index to multiplied variables.

$$a^{m}\times b^{m}=\left(ab \right )^{m}$$

For example:

$$3^{2}\times 5^{2}=\left(3\times 5 \right )^{2}$$

$$\left(15 \right )^{2}=225$$

Rule 7: When two variables with different bases but the same indices are divided.

Then, $$\frac{a^{m}}{b^{m}}=\left(\frac{a}{b} \right )^{m}$$

For example$$\frac{3^{2}}{5^{2}}=\left(\frac{3}{5} \right )^{2}=\frac{9}{25}$$

Rule 8: An index in the form of a fraction can be represented as the radical form.

$$a^{\frac{m}{n}}=\sqrt[n]{a^{m}}$$

For example$$6^{\frac{1}{2}}=\sqrt{6}$$

### Worked examples

Example 1:Solve $$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}$$.

Step 1: Write the given equation.

$$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}$$.

Step 2: Apply the rule $$\frac{a^{m}}{b^{m}}=\left(\frac{a}{b} \right )^{m}$$ in the given equation.

$$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}=\frac{2^{2}\left(a^{4} \right )^{2}b^{2}\times a^{3}\left(b^{2} \right )^{3}}{6^{2}\left(a^{2} \right )^{2}\left(b^{3} \right )^{2}}$$

Step 3: Apply the rule $$\left(a^{m} \right )^{n}=a^{mn}$$ in the above equation.

The above equation can be written as follows:

$$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}=\frac{4a^{8}b^{2}\times a^{3}b^{6}}{36a^{4}b^{6}}$$

Step 4: Apply the rule $$a^{m}\times a^{n}=a^{m+n}$$ in the above equation.

$$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}=\frac{4a^{(8+3)}b^{(2+6)}}{36a^{4}b^{6}}$$

$$\frac{4a^{11}b^{8}}{36a^{4}b^{6}}$$

Step 5: Apply the rule $$\frac{a^{m}}{a^{n}}=a^{m-n}$$ in the above equation.

$$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}=\frac{1}{9}a^{(11-4)}b^{(8-6)}$$

$$\frac{1}{9}a^{7}b^{2}$$

So, the solution is $$\frac{\left(2a^{4}b \right )^{2}\times \left(ab^{2} \right )^{3}}{\left(6a^{2}b^{3} \right )^{2}}=\frac{1}{9}a^{7}b^{2}$$

Example 2: Solve $$9^{\frac{5}{2}}-3-\left(\frac{1}{81} \right )^{\frac{-1}{2}}$$.

Step 1: Write the given equation.

$$9^{\frac{5}{2}}-3-\left(\frac{1}{81} \right )^{\frac{-1}{2}}$$.

Step 2: Rewrite the given equation.

$$9^{\frac{5}{2}}-3-\left(\frac{1}{81} \right )^{\frac{-1}{2}}=\left(3^{2} \right )^{\frac{5}{2}}-3-\left(\frac{1}{3^{4}} \right )^{\frac{-1}{2}}$$

Step 3: Apply the rule $$\left(a^{m} \right )^{n}=a^{mn}$$ and $$a^{-m}=\frac{1}{a^{m}}$$ in the above equation.

$$9^{\frac{5}{2}}-3-\left(\frac{1}{81} \right )^{\frac{-1}{2}}=3^{5}-3-3^{2}$$

$$243-3-9=231$$

So, the solution is $$9^{\frac{5}{2}}-3-\left(\frac{1}{81} \right )^{\frac{-1}{2}}=231$$