Laws of Exponents

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Laws of Exponents

The Laws of Exponents, also known as the Rules of Exponents, are a set of rules that describe how to handle mathematical operations involving exponents or powers. These laws apply when working with powers of the same base. Below are the main laws:

a^m × aⁿ = a^m+n

(a^m)ⁿ = a^mn

(a × b)^m = a^m × b^m or (ab)^m = a^m × b^m

a⁰ =1, (a ≠ 0)

a^m ÷ aⁿ = a^{m – n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

$a^{- m} = \frac{1}{a ^{m}}$

Note :

$(a^{m})^{n} \neq = a^{mn}$

E.g.,

$(2^{3})^{2} = 2^{3 \times 2} = 2^{6} = 64 and 2^{3^{2}} = 2^{9} = 512$

1.0Scientific Notation

The Sun is located at a distance of 1,000,000,000,000,000,000 km from the center of milky way. The average size of an atom is about 0.00000003 centimetre across.

The length of these numbers in standard notation makes them awkward to work with. Scientific notation is a shorthand way of writing such numbers.

To express any number in scientific notation, write it as the product of a power of ten and a number greater than or equal to 1 but less than 10.

In scientific notation, the Sun is located at a distance of 1.0 × 10¹⁸ m from the centre of milky way and the size of each atom is 3.0 × 10^–8 centimetres across.

2.0Astronomy Calculation

The distance between the Sun and the Moonis 1.49984 × 10¹¹ m and the distance between the Moon and the Earth is 3.84 × 10⁸ m. During the Lunar eclipse, the Earth comes between the Sun and the Moon. The distance between the Sun and the Earth at this time.

Distance between:

The Sun and the Moon = 1.49984 × 10¹¹ m

The Moon and the Earth = 3.84 × 10⁸ m

The Sun and the Earth = 1.49984 × 10¹¹ – 3.84 × 10⁸

= 1.49984 × 1000 × 10⁸ – 3.84 × 10⁸

= 1499.84 × 10⁸ – 3.84 × 10⁸

= (1499.84 – 3.84) × 10⁸

= 1496 × 10⁸

= 1.496 × 10¹¹ m

3.0Solved Examples

1.Simplify and write the answer in an exponential form.

(i) (3⁷ ÷ 3¹¹)³ × 3^–8

(ii) $[(\frac{6}{7})^{3}]^{2} \times [(\frac{6}{7})^{- 4}]^{2}$

Solution

(i) (3⁷ ÷ 3¹¹)³ × 3^–8

= (3^7–11)³ × 3^–8= (3^–4)³ × 3^–8 = 3^–12 × 3^–8 = 3^{–12 + (–8)}

= 3^–20= $\frac{1}{3 ^{20}}$

(ii) $[(\frac{6}{7})^{3}]^{2} \times [(\frac{6}{7})^{- 4}]^{2} = (\frac{6}{7})^{3 \times 2} \times (\frac{6}{7})^{- 4 \times 2}$

$= (\frac{6}{7})^{6} \times (\frac{6}{7})^{- 8} = (\frac{6}{7})^{6 + (- 8)} = (\frac{6}{7})^{- 2} = (\frac{7}{6})^{2}$

2.Find the value of x.

$(\frac{- 11}{7})^{- 5} \times (\frac{- 11}{7})^{x} = (\frac{- 11}{7})^{3}$

Solution

$(\frac{- 11}{7})^{- 5} \times (\frac{- 11}{7})^{x} = (\frac{- 11}{7})^{3}$

$(\frac{- 11}{7})^{(- 5) + x} = (\frac{- 11}{7})^{3}$

Since, the base is same on both sides of the expression, their exponents should also be the same.

⇒ –5 + x = 3 ⇒ x = 3 + 5

∴ x = 8

3.Find the value of the following using laws of exponents.

(i) $\frac{81 \times 7 ^{3} \times 100}{1 0 ^{2} \times 3 ^{4} \times 7}$

(ii) $\frac{2 ^{8} \times 3 ^{4}}{8 \times ( - 2 ) ^{5} \times 27}$

Solution

(i) $\frac{3 ^{4} \times 7 ^{3} \times 1 0 ^{2}}{1 0 ^{2} \times 3 ^{4} \times 7} = 3^{4 - 4} \times 7^{3 - 1} \times 1 0^{2 - 2} = 3^{0} \times 7^{2} \times 1 0^{0}$

= 1 × 49 × 1 = 49

(ii) $\frac{2 ^{8} \times 3 ^{4}}{2 ^{3} \times ( - 1 ) ^{5} \times ( 2 ^{5} ) \times 3 ^{3}} = \frac{2 ^{8 - 3 - 5} \times 3 ^{4 - 3}}{( - 1 )}$

= –1 × 2⁰ × 3 = – 1 × 3 = – 3 (2⁰ = 1)

4.By what number should $(\frac{- 2}{3})^{- 3}$ be divided so that the quotient may be $(\frac{4}{27})^{- 2}$ ?

Solution

$(\frac{- 2}{3})^{- 3} \div X = (\frac{4}{27})^{- 2}$

$X = (\frac{- 2}{3})^{- 3} \div (\frac{4}{27})^{- 2} = (\frac{- 2}{3})^{- 3} \times (\frac{2 ^{2}}{3 ^{3}})^{2} = (- 1^{- 3}) (\frac{2 ^{- 3 + 4}}{3 ^{- 3 + 6}}) = \frac{- 2}{3 ^{3}} = \frac{- 2}{27}$

5.By what number should (4)^–3be multiplied so that the product become ?

Solution

$(4)^{- 3} \times Z = \frac{1}{16} ⟹ Z = \frac{1}{4 ^{2}} \div (4)^{- 3}$

$Z = \frac{1}{4 ^{2}} \div \frac{1}{4 ^{3}}$ ⇒ $\frac{1}{4 ^{2}} \times 4^{3} = 4^{3 - 2} = 4$

6.Write the numbers in the standard form.

(i) 0.00925

(ii) 457000000

(iii) 0.32458

Solution

(i) 0.00925 = 9.25 × 10^–3

(ii) 457000000 = 4.57 × 10⁸

(iii) 0.32458 = 3.2458 × 10^–1

7. Express the following in scientific notation.

(i) The weight of the earth is about 6,600,000,000,000,000,000,000 metric tons.

(ii) The speed of light is 300,000,000 metres per second.

(iii) The distance between the earth and moon is 382,000 km.

(iv) The surface area of the earth is 510,200,000 km².

Solution

(i) 6.6 × 10²¹metric tons

(ii) 3 × 10⁸meters per second

(iii) 3.82 × 10⁵km

(iv) 5.102 × 10⁸km²

8.The thickness of a sheet of paper is 1.6 × 10^–3cm and the thickness of a human hair is 5 × 10^–3 cm. Compare them.

Solution

$\frac{Thickness of hair}{Thickness of paper} = \frac{5 \times 1 0 ^{- 3}}{1.6 \times 1 0 ^{- 3}} = \frac{5 \times 1 0 ^{- 3} \times 1 0 ^{3}}{1.6} = \frac{5}{1.6} = 3.125$

The hair is approximately three times thicker than the paper.

4.0Also Read

Standard Units of Measurement	Circles	Introduction to Numbers
Ascending Order	Area of Rectangle	Roman Numerals
Area of Irregular Shapes	Area of A Circle	Rounding Numbers

Laws of Exponents

The Laws of Exponents, also known as the Rules of Exponents, are a set of rules that describe how to handle mathematical operations involving exponents or powers. These laws apply when working with powers of the same base. Below are the main laws:

a^m × aⁿ = a^m+n

(a^m)ⁿ = a^mn

(a × b)^m = a^m × b^m or (ab)^m = a^m × b^m

a⁰ =1, (a ≠ 0)

a^m ÷ aⁿ = a^{m – n}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

$a^{- m} = \frac{1}{a ^{m}}$

Note :

$(a^{m})^{n} \neq = a^{mn}$

E.g.,

$(2^{3})^{2} = 2^{3 \times 2} = 2^{6} = 64 and 2^{3^{2}} = 2^{9} = 512$

1.0Scientific Notation

The Sun is located at a distance of 1,000,000,000,000,000,000 km from the center of milky way. The average size of an atom is about 0.00000003 centimetre across.

The length of these numbers in standard notation makes them awkward to work with. Scientific notation is a shorthand way of writing such numbers.

To express any number in scientific notation, write it as the product of a power of ten and a number greater than or equal to 1 but less than 10.

In scientific notation, the Sun is located at a distance of 1.0 × 10¹⁸ m from the centre of milky way and the size of each atom is 3.0 × 10^–8 centimetres across.

2.0Astronomy Calculation

The distance between the Sun and the Moonis 1.49984 × 10¹¹ m and the distance between the Moon and the Earth is 3.84 × 10⁸ m. During the Lunar eclipse, the Earth comes between the Sun and the Moon. The distance between the Sun and the Earth at this time.

Distance between:

The Sun and the Moon = 1.49984 × 10¹¹ m

The Moon and the Earth = 3.84 × 10⁸ m

The Sun and the Earth = 1.49984 × 10¹¹ – 3.84 × 10⁸

= 1.49984 × 1000 × 10⁸ – 3.84 × 10⁸

= 1499.84 × 10⁸ – 3.84 × 10⁸

= (1499.84 – 3.84) × 10⁸

= 1496 × 10⁸

= 1.496 × 10¹¹ m

3.0Solved Examples

1.Simplify and write the answer in an exponential form.

(i) (3⁷ ÷ 3¹¹)³ × 3^–8

(ii) $[(\frac{6}{7})^{3}]^{2} \times [(\frac{6}{7})^{- 4}]^{2}$

Solution

(i) (3⁷ ÷ 3¹¹)³ × 3^–8

= (3^7–11)³ × 3^–8= (3^–4)³ × 3^–8 = 3^–12 × 3^–8 = 3^{–12 + (–8)}

= 3^–20= $\frac{1}{3 ^{20}}$

(ii) $[(\frac{6}{7})^{3}]^{2} \times [(\frac{6}{7})^{- 4}]^{2} = (\frac{6}{7})^{3 \times 2} \times (\frac{6}{7})^{- 4 \times 2}$

$= (\frac{6}{7})^{6} \times (\frac{6}{7})^{- 8} = (\frac{6}{7})^{6 + (- 8)} = (\frac{6}{7})^{- 2} = (\frac{7}{6})^{2}$

2.Find the value of x.

$(\frac{- 11}{7})^{- 5} \times (\frac{- 11}{7})^{x} = (\frac{- 11}{7})^{3}$

Solution

$(\frac{- 11}{7})^{- 5} \times (\frac{- 11}{7})^{x} = (\frac{- 11}{7})^{3}$

$(\frac{- 11}{7})^{(- 5) + x} = (\frac{- 11}{7})^{3}$

Since, the base is same on both sides of the expression, their exponents should also be the same.

⇒ –5 + x = 3 ⇒ x = 3 + 5

∴ x = 8

3.Find the value of the following using laws of exponents.

(i) $\frac{81 \times 7 ^{3} \times 100}{1 0 ^{2} \times 3 ^{4} \times 7}$

(ii) $\frac{2 ^{8} \times 3 ^{4}}{8 \times ( - 2 ) ^{5} \times 27}$

Solution

(i) $\frac{3 ^{4} \times 7 ^{3} \times 1 0 ^{2}}{1 0 ^{2} \times 3 ^{4} \times 7} = 3^{4 - 4} \times 7^{3 - 1} \times 1 0^{2 - 2} = 3^{0} \times 7^{2} \times 1 0^{0}$

= 1 × 49 × 1 = 49

(ii) $\frac{2 ^{8} \times 3 ^{4}}{2 ^{3} \times ( - 1 ) ^{5} \times ( 2 ^{5} ) \times 3 ^{3}} = \frac{2 ^{8 - 3 - 5} \times 3 ^{4 - 3}}{( - 1 )}$

= –1 × 2⁰ × 3 = – 1 × 3 = – 3 (2⁰ = 1)

4.By what number should $(\frac{- 2}{3})^{- 3}$ be divided so that the quotient may be $(\frac{4}{27})^{- 2}$ ?

Solution

$(\frac{- 2}{3})^{- 3} \div X = (\frac{4}{27})^{- 2}$

$X = (\frac{- 2}{3})^{- 3} \div (\frac{4}{27})^{- 2} = (\frac{- 2}{3})^{- 3} \times (\frac{2 ^{2}}{3 ^{3}})^{2} = (- 1^{- 3}) (\frac{2 ^{- 3 + 4}}{3 ^{- 3 + 6}}) = \frac{- 2}{3 ^{3}} = \frac{- 2}{27}$

5.By what number should (4)^–3be multiplied so that the product become ?

Solution

$(4)^{- 3} \times Z = \frac{1}{16} ⟹ Z = \frac{1}{4 ^{2}} \div (4)^{- 3}$

$Z = \frac{1}{4 ^{2}} \div \frac{1}{4 ^{3}}$ ⇒ $\frac{1}{4 ^{2}} \times 4^{3} = 4^{3 - 2} = 4$

6.Write the numbers in the standard form.

(i) 0.00925

(ii) 457000000

(iii) 0.32458

Solution

(i) 0.00925 = 9.25 × 10^–3

(ii) 457000000 = 4.57 × 10⁸

(iii) 0.32458 = 3.2458 × 10^–1

7. Express the following in scientific notation.

(i) The weight of the earth is about 6,600,000,000,000,000,000,000 metric tons.

(ii) The speed of light is 300,000,000 metres per second.

(iii) The distance between the earth and moon is 382,000 km.

(iv) The surface area of the earth is 510,200,000 km².

Solution

(i) 6.6 × 10²¹metric tons

(ii) 3 × 10⁸meters per second

(iii) 3.82 × 10⁵km

(iv) 5.102 × 10⁸km²

8.The thickness of a sheet of paper is 1.6 × 10^–3cm and the thickness of a human hair is 5 × 10^–3 cm. Compare them.

Solution

$\frac{Thickness of hair}{Thickness of paper} = \frac{5 \times 1 0 ^{- 3}}{1.6 \times 1 0 ^{- 3}} = \frac{5 \times 1 0 ^{- 3} \times 1 0 ^{3}}{1.6} = \frac{5}{1.6} = 3.125$

The hair is approximately three times thicker than the paper.

4.0Also Read

Standard Units of Measurement	Circles	Introduction to Numbers
Ascending Order	Area of Rectangle	Roman Numerals
Area of Irregular Shapes	Area of A Circle	Rounding Numbers

Join ALLEN!

Join ALLEN!

Laws of Exponents

1.0Scientific Notation

2.0Astronomy Calculation

3.0Solved Examples

4.0Also Read

Table of Contents

Laws of Exponents

1.0Scientific Notation

2.0Astronomy Calculation

3.0Solved Examples

4.0Also Read

Table of Contents