# Laws Of Exponents With Integral Power

Exponents are used to show repeated multiplication of a number by itself. For example, 7 × 7 × 7 can be represented as $7^3$. In this example the exponent term is ‘3’ which stands for the number of times the value is multiplied. Here 7 is called the base which is the actual number that is getting multiplied. Exponents follow certain rules that help in simplifying expressions known as the laws of exponents. Let us discuss the laws of exponents in detail.

##### Laws of exponents

1. For any non-zero term $a$, $am~×~an$ = $am + n$ where $m$ and $n$ are natural numbers.

Illustration 1: What is the simplification of $5^5~×~5^1$?

Solution: $5^5~×~5^1$ = $5^{5+1}$ = $5^6$.

Illustration 2: What is the simplification of $(-6)^{-4}~×~(-6)^{-7}$?

Solution: (-6)^(-4) = – $\frac{1}{6^4}$ and $(-6)^{-7}$ = $-{1}{6^7}$

$(-6)^{-4}~×~(-6)^{-7}$ = $-\frac{1}{6^4}~×~-\frac{1}{6^7}$ = $+ \frac{1}{6^{4+7}}$ = ${1}{6^11}$ = $6^{-11}$.

We can state that the law is applicable for negative terms also. Therefore the term m and n can be any integer.

2. $\frac{a^m}{a^n}$ = a^(m-n),where $a$ is a non-zero term and $m$ and $n$ are integers.

Illustration 3: Find the value when $10^{-5}$ is divided by $10^{-3}$.

Solution: $\frac{10^{-5}}{10^{-3}}$ = $10^{-5-(-3)}$ = $10^{-5+3}$ = $10^{-2}$ = $\frac{1}{100}$

3. $(a^m)^n$ = a^(mn), where a is a non-zero term and $m$ and $n$ are integers.

Illustration 4: Express $8^3$ as a power with base 2.

Solution: We have, $2×2×2$ = $8$ = 2^3

Therefore, $8^3$ = $(2^3)^3$ = 2^9.

4. a^m × b^m =(ab)^m, where a is a non-zero term and m and n are integers.

Illustration 5: Simplify and write the exponential form of: $\frac{1}{8}~ ×~ 5^{-3}$.

Solution: We know that $\frac{1}{8}$ = 2^{-3}

Therefore, $2^{-3}~×~5^{-3}$ = $(2 × 5)^{-3}$ = $10^{-3}$.

5. $\frac{a^m}{b^m}$ = $(\frac{a}{b})^m$, where $a$ and $b$ are non-zero terms and $m$ is an integer.

Illustration 6: Simplify the expression and find the value: $\frac{15^3}{5^3}$ .

Solution: $(\frac{15}{5})^3$ = $(3)^3$ = $27$

6. $a^0$ = $1$, where $a$ is any non-zero term.

Illustration 7: What is the value of $5^0 + 2^2 + 4^0 + 7^1 – 3^1$ ?

Solution: $5^0 + 2^2 + 4^0 + 7^1 – 3^1$ = $1 + 4 + 1 + 7 – 3$ = $10$<

This article covers the basic laws of exponents. For any further query on this topic please install Byju’s the learning app.

#### Practise This Question

Which of the following fractions is equal to 157?