 # Exponents and Powers Class 7

Exponents and Power class 7 topic includes the basic concepts of exponents and powers. You will be learning the exponent properties, formulas, scientific notation, orders of magnitude, laws of exponents and other related fundamentals in this chapter. The questions of exponents and powers can be easily solved once you cover all the topics and understand them briefly.

## Exponents and Powers Class 7 Topics

The topics and sub-topics covered in this chapter include:

• Introduction
• Exponents
• Laws of exponents
• Multiplying Powers With the Same Base
• Dividing Powers with the Same Base
• Taking Power of a Power
• Multiplying Powers with the Same Exponents
• Dividing Powers with the Same Exponents
• Miscellaneous Examples Using the Laws of Exponents
• Decimal Number System
• Expressing Large Numbers using Standard Form

## Exponents and Powers Class 7 Notes

The exponents state the repeated multiplication of the same number with respect to the number of times. For example, 23 = 2 x 2 x 2. Therefore, using exponents means raising a number to a power, where the exponent is the power. The number to which the power is raised is called the base of the power, thus, in 23, 2 is the base and 3 is the exponent. The difference between the power and exponents are, the exponent is the power raised to the base number and power is represented as the combination of base number and exponent. Thus, 23 represents power.

To denote the exponent value, the base number value should be the same and then altogether it is said to be a power. Exponents and Powers are used to express large numbers which cannot be represented in the general form, such as 100000000000000 can be represented as 1014. In this article, apart from basics, you will also learn exponents and powers class 7 formulas with some set of examples.

## Exponents and Powers Class 7 Formulas

If p is a rational number and have a non-zero value, m is a natural number, then,

p × p × p × p ×…..× p(m times) is written as pm, where p is the base number and m is the exponent value and pm is the power and ‘pm’ is said as ‘p – raised to the power m’. This is the general representation of exponents and powers.

Example: 9 × 9 × 9 × 9 × 9 × 9 × 9 = 97, where 9 is the base number and 7 is the exponent.

There are key to laws of exponents defined to solve complex problems based on powers and exponents.

### Laws of Exponents

If p and q are non- zero rational numbers and m and n are natural numbers, then the laws of exponents can be written as;

• pm x pn = pm+n
• (pm)n = pmn
• $$\begin{array}{l}\frac{p^m}{p^n}\end{array}$$
= pm-n ; where m>n
• $$\begin{array}{l}\frac{p^m}{p^n}\end{array}$$
= 1/pn-m ; where n>m
• (pq)m = pm qm
• $$\begin{array}{l}(\frac{p}{q})^m\end{array}$$
=
$$\begin{array}{l}\frac{p^m}{q^m}\end{array}$$

Let us solve some problems based on these laws of exponents to understand the concepts clearly.

### Exponent and Powers Class 7 Example Problems

Problem 1:

Solve: 23 x 22

Solution:

From the law of exponent we know,

pm x pn = pm+n

Therefore, 23 x 22 = 23+2 = 25 = 2 x 2 x 2 x 2 x 2 = 32

Problem 2:

Solve:

$$\begin{array}{l}\frac{3^3}{3^2}\end{array}$$

Solution:

Applying the law of exponent, we get,

$$\begin{array}{l}\frac{p^m}{p^n}\end{array}$$
= pm-n

So,

$$\begin{array}{l}\frac{3^3}{3^2}\end{array}$$
= 33-2 = 31 = 3

Problem 3:

Solve: (52)2

Solution:

We know, by the law of exponent,

(pm)n = pmn

Therefore, (52)2 = 52×2 = 54 = 625

Problem 4:

$$\begin{array}{l}(\frac{10}{5})^2\end{array}$$
?

Solution:

By the law of exponent, we can write the given equation as;

$$\begin{array}{l}(\frac{10}{5})^2\end{array}$$
=
$$\begin{array}{l}\frac{10^2}{5^2}\end{array}$$

=

$$\begin{array}{l}\frac{100}{25}\end{array}$$

= 4

Problem 5:

Find the value of

$$\begin{array}{l}\frac{2^3}{2^5}\end{array}$$
.

Solution:

From the given problem, we can see the power of the denominator is greater than the power in the numerator.

Therefore, by the law of exponent,

$$\begin{array}{l}\frac{p^m}{p^n}\end{array}$$
= 1/pn-m ; where n>m

$$\begin{array}{l}\frac{2^3}{2^5}\end{array}$$
= 1/25-3

= 1/22

= ¼