Exponents

Table of Contents

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The exponent of a number shows how many times the number is multiplied by itself.

For example, \(2 \times 2 \times  2 \times  2 \) can be written as \(2^4\); as \(2\) is multiplied by itself \(4\) times.

Here, \(2\) is called the "base" and \(4\) is called the "exponent" or "power."

In general, \(x^n\) means that \(x\) is multiplied by itself for \(n\) times.

Here,

• \(x\) is called the "base"
• \(n\) is called the "exponent" or "power"
• \(x^n\) is read as "\(x\) to the power of \(n\)" (or) "\(x\) raised to \(n\)"

Examples of Exponents

Some examples of exponents are as follows:

• \(\begin{align} 3 \times 3 \times 3 \times 3 \times 3 &= 3^5\end{align}\)
   
• \(\begin{align}-2 \times -2 \times -2 &= (-2)^3\end{align}\)
   
• \(\begin{align}a \times a \times a \times a \times a \times a &=a^6\end{align}\)

Why are Exponents Important?

Exponents are important because, without them, the products where a number is repeated by itself for many times is very difficult to write.

For example, it is very easy to write \(5^7\) instead of writing \(5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5\).

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Here is the "Exponents Calculator".

Here, you can set the values of the base and the exponent.

Then it will show the corresponding value.

The laws (properties or rules) of exponents are used to solve problems involving exponents.

The laws of exponents are mentioned below.

For each law, an example is provided to see the difference in solving equations when we use laws of exponents versus when we do not use the laws.

Law of Product:

\(\mathbf{a^m \times a^n = a^{m + n}}\)

Using the law	Without using the law
\(\begin{align}2^3 \times 2^5 &= 2^{3+5}\\&=2^8 \end{align}\)	\(\begin{aligned} 2^{3} \times 2^{5} &=(2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2) \\ &=2^{8} \end{aligned}\)

Law of Quotient:

\(\mathbf{\dfrac{a^m}{a^n}= a^{m-n}} \)

Using the law	Without using the law
\(\begin{align}\dfrac{2^5}{2^3}&= 2^{5-3}\\&=2^2\end{align}\)	\( \begin{aligned} \frac{2^{5}}{2^{3}} &=\frac{2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2} \\ &=2^{2} \end{aligned} \)

Law of Zero Exponent:

\(\mathbf{a^0=1} \)

Using the law	Without using the law
\(2^0 =1\)	\begin{aligned}2^{0} &=2^{5-5} \\&=\frac{2^{5}}{2^{5}}\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right] \\&=\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} \\&=1\end{aligned}

Law of Negative Exponent:

\(\mathbf{a^{-m} = \dfrac{1}{a^m}} \)

Using the law	Without using the law
\(2^{-2} = \dfrac{1}{2^2}\)	\(\begin{aligned} 2^{-2} &=2^{0-2} \\ &=\frac{2^{0}}{2^{2}}\left[\because \frac{a^{m}}{a^{n}}=a^{m-n}\right] \\ &=\frac{1}{2^{2}}\left[\because 2^{0}=1\right] \end{aligned} \)

Law of Power of a Power:

\(\mathbf{(a^m)^n=a^{mn}}\)

Using the law	Without using the law
\((2^2)^3 = 2^6\)	\begin{aligned} \left(2^{2}\right)^{3} &=\left(2^{2}\right)\left(2^{2}\right)\left(2^{2}\right) \\ &=(2 \cdot 2)(2 \cdot 2)(2 \cdot 2) \\ &=2^{6} \end{aligned}

Law of Power of a Product:

\(\mathbf{(ab)^m=a^mb^m}\)

Using the law	Without using the law
\((xy)^3 = x^3y^3\)	\( \begin{aligned} (x y)^{3} &=(x y)(x y)(x y) \\ &=(x x x)(y y y) \\ &=x^{3} y^{3} \end{aligned} \)

Law of Power of a Quotient:

\(\mathbf{\left(\dfrac{a}{b}\right)^m= \dfrac{a^m}{b^m}}\)

Using the law	Without using the law
\(\left(\dfrac{x}{y}\right)^3= \dfrac{x^3}{y^3}\)	\( \begin{aligned} \left(\frac{x}{y}\right)^{3} &=\frac{x}{y} \cdot \frac{x}{y} \cdot \frac{x}{y} \\ &=\frac{x^{3}}{y^{3}} \end{aligned} \)

From these tables above, you would have got an idea of how and why a particular law works and how these laws of exponents make our work easier.

Let's try and solve a few examples using these laws.

Example 1

 

 

The dimensions of a wardrobe are \(x^5\) cm, \(y^3\) cm and \(x^8\) cm. Find its volume.

Solution:

The dimensions of the wardrobe are:

\[ \begin{aligned} \text{Length, } l &= x^5 \text{ cm}\\[0.2cm] \text{Width, } w &= y^3 \text{ cm}\\[0.2cm] \text{Height, } h &= x^8 \text{ cm} \end{aligned} \]

The volume of the wardrobe, \(V\) is:

\[ \begin{aligned} V &= lwh\\[0.2cm]&=x^5 \cdot y^3 \cdot x^8\\[0.2cm]&=x^{13} \,y^3 \; [\because a^m \cdot a^n = a^{m+n} ] \end{aligned} \]

\( \therefore V =x^{13} \,y^3 \)

Example 2

 

 

Evaluate the following expression as an integer:

\[\left(\dfrac{1}{2}\right)^{-2}+\left(\dfrac{1}{3}\right)^{-2}+\left(\dfrac{1}{5}\right)^{-2}\]

Solution:

We will simplify the given expression using the laws of exponents.

\(\begin{align}
&= \frac{1^{-2}}{2^{-2}} + \frac{1^{-2}}{3^{-2}} + \frac{1^{-2}}{5^{-2}} \;[\because \left(\dfrac{a}{b}\right)^m\!= \dfrac{a^m}{b^m}] \\\\
&= \frac{1}{2^{-2}}+\frac{1}{3^{-2}}+\frac{1}{5^{-2}} \;[\because 1^{-2}=1] \\\\
&= 2^2+3^2+5^2 \; [\because \dfrac{1}{a^{-m}} = a^m]\\\\
&= 4+9+25\\\\
&=38
\end{align}\)

\(\therefore\) \(\begin{align}\left(\frac{1}{2}\right)^{-2}\!\!+\!\left(\frac{1}{3}\right)^{-2}\!\!+\!\left(\frac{1}{5}\right)^{-2} \!= 38\end{align}\)

Example 3

 

 

Simplify the following expression:

\[ \dfrac{\left(x^{a+b}\right)^{2} \times\left(x^{b+c}\right)^{2} \times\left(x^{c+a}\right)^{2}}{\left(x^{a} \times x^{b} \times x^{c}\right)^{3}} \]

Solution:

We will simplify the given expression using the laws of exponents.

\begin{align}
&\frac{\left(x^{a+b}\right)^{2} \times\left(x^{b+c}\right)^{2} \times\left(x^{c+a}\right)^{2}}{\left(x^{a} \cdot x^{b} \cdot x^{c}\right)^{3}}\\\\
&= \frac{x^{2a+2b}\times x^{2b+2c}\times x^{2c+2a}}{x^{3a} \times x^{3b} \times x^{3c}} \;[\because \! (a^m)^n \!\!=\! a^{mn}; (ab)^m \!\!=\! a^m b^m ]\\\\ &=\frac{x^{2a+2b+2b+2c+2c+2a}}{x^{3a+3b+3c}} \;[\because \!a^m \!\!\times\! a^n \!\!=\!a^{m+n} ]\\\\ &= \frac{x^{4a+4b+4c}}{x^{3a+3b+3c}}\\\\ &=x^{4a+4b+4c - (3a+3b+3c)} \; [\because \!\dfrac{a^m}{a^n}\!\!=\! a^{m-n} ]\\\\&= x^{a+b+c}
\end{align}

\(\begin{align}\therefore\!\frac{\left(x^{a+b}\right)^{2} \!\times\!\left(x^{b+c}\right)^{2} \!\times\!\left(x^{c+a}\right)^{2}}{\left(x^{a} \!\cdot\! x^{b} \!\cdot\! x^{c}\right)^{3}} \!= x^{a+b+c}\end{align}\)

Example 4

 

 

In a forest, each tree has about \(5^7\) leaves and there are about \(5^3\) trees in the forest.

Find the total number of leaves.

Solution:

The number of trees in the forest = \(5^3\)

The number of leaves in each tree = \(5^7\)

The total number of leaves are:

\[ \begin{aligned}  5^3 \times 5^7 &= 5^{10} \; [\because a^m \times a^n = a^{m+n}] \end{aligned} \]

\(\therefore\) Total number of leaves \(=5^{10}\)

Example 5

 

 

Find the value of \(x\) if:

\[ \left(\frac{1}{7}\right)^{2} \times\left(\frac{1}{7}\right)^{-9}=\left(\frac{1}{7}\right)^{2 x-3} \]

Solution:

The given equation is:

\[ \begin{aligned} \left(\frac{1}{7}\right)^{2} \!\!\!\times\!\left(\frac{1}{7}\right)^{-9}&\!\!\!=\!\left(\frac{1}{7}\right)^{2 x-3} \\[0.3cm] \left(\frac{1}{7}\right)^{-7} &\!\!\!=\! \left(\frac{1}{7}\right)^{2x-3} \; [\because \!a^m \!\times\! a^n\!=\!a^{m+n} ] \\[0.3cm] -7 &\!\!=\! 2x-3 \; [\because\! \text{The bases are equal} ]\\[0.3cm] -4 &\!\!=\!2x \\[0.3cm] -2 &\!\!=\! x \end{aligned} \]

\( \therefore x = -2 \)

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