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.

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

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

The laws of exponents are mentioned in the table below.

For each law, an example is provided which is solved using the law.

Law of Exponents	Examples using the law
\(a^m \times a^n = a^{m + n}\)	\(\begin{align}x^3 \times x^5 &= x^{3+5}\\&=x^8 \end{align}\)
\(\dfrac{a^m}{a^n}= a^{m-n}\)	\(\begin{align}\dfrac{x^5}{x^3}&= x^{5-3}\\&=x^2\end{align}\)
\(a^0=1\)	\(x^0 =1\)
\(a^{-m} = \dfrac{1}{a^m} \)	\(x^{-2} = \dfrac{1}{x^2}\)
\((a^m)^n=a^{mn}\)	\((x^2)^3 = x^6\)
\((ab)^m=a^mb^m\)	\((xy)^3 = x^3y^3\)
\(\left(\dfrac{a}{b}\right)^m= \dfrac{a^m}{b^m}\)	\(\left(\dfrac{x}{y}\right)^3= \dfrac{x^3}{y^3}\)

From this table 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 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:

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

\( \therefore x = -2 \)