Exponents

Definition of Exponents

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.

Definition of exponents: x raised to n is x 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 \(n\)"

Definition of exponents: In x raised to n, x is the base and n is the exponent

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 \(5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5\).


Exponents Calculator

Here is the "Exponents Calculator".

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

Then it will show the corresponding value.


Laws (Properties or Rules) of Exponents

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.

 
important notes to remember
Important Notes

The laws of exponents are:
\begin{aligned} a^0&=1\\[0.2cm]a^1&=a\\[0.2cm]a^m \times a^n &= a^{m + n}\\[0.2cm]\dfrac{a^m}{a^n}&= a^{m-n}\\[0.2cm]a^{-m} &= \dfrac{1}{a^m}\\[0.2cm](a^m)^n&=a^{mn}\\[0.2cm](ab)^m&=a^mb^m\\[0.2cm]\left(\dfrac{a}{b}\right)^m&= \dfrac{a^m}{b^m}\end{aligned}

Solved Examples

Example 1

 

 

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

Solved examples with exponents - The volume of a wardrobe is to be calculated. The length, width, and height are given in exponents.

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.

Solved problems for exponents - The total number of leaves in a forest is to be calculated.

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:

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

\( \therefore x = -2 \)

Practice Questions

 
 
 
 
 
 
 
tips and tricks
Tips and Tricks
  1. If a fraction has a negative exponent, then we take the reciprocal of the fraction to make the exponent positive. i.e.,
    \[ \left(\dfrac{a}{b} \right)^{-m}= \left(\dfrac{b}{a} \right)^{m} \]
  2. When the exponents are the same, we can set the bases equal and vice versa. i.e.,
    \[ a^m = a^n \Leftrightarrow m =n \]

Exponents Worksheets

You can download the exponents worksheets available at the bottom of this page.

You can work on them and get expertised in this topic.


Maths Olympiad Sample Papers

IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. It encourages children to develop their math solving skills from a competition perspective.

You can download the FREE grade-wise sample papers from below:

To know more about the Maths Olympiad you can click here


Frequently Asked Questions (FAQs)

1. What are the Laws of Exponents?

The laws of exponents are:

\begin{aligned} a^0&=1\\[0.2cm]a^1&=a\\[0.2cm]a^m \times a^n &= a^{m + n}\\[0.2cm]\dfrac{a^m}{a^n}&= a^{m-n}\\[0.2cm]a^{-m} &= \dfrac{1}{a^m}\\[0.2cm](a^m)^n&=a^{mn}\\[0.2cm](ab)^m&=a^mb^m\\[0.2cm]\left(\dfrac{a}{b}\right)^m&= \dfrac{a^m}{b^m}\end{aligned}

2. What are the examples of exponents?

Some examples of exponents are as follows:

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

3. What are the rules of exponents?

The rules of exponents are:

\begin{aligned} a^0&=1\\[0.2cm]a^1&=a\\[0.2cm]a^m \times a^n &= a^{m + n}\\[0.2cm]\dfrac{a^m}{a^n}&= a^{m-n}\\[0.2cm]a^{-m} &= \dfrac{1}{a^m}\\[0.2cm](a^m)^n&=a^{mn}\\[0.2cm](ab)^m&=a^mb^m\\[0.2cm]\left(\dfrac{a}{b}\right)^m&= \dfrac{a^m}{b^m}\end{aligned}

Download Exponents and Logarithms Worksheets
Exponents and Logarithms
Grade 9 | Answers Set 1
Exponents and Logarithms
Grade 9 | Answers Set 2
Exponents and Logarithms
Grade 9 | Questions Set 1
  
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