Laws of Exponents

The laws of exponents are explained here along with their examples.

1. Multiplying Powers with same Base

For example: x² × x³, 2³ × 2⁵, (-3)² × (-3)⁴

In multiplication of exponents if the bases are same then we need to add the exponents.

Consider the following: 

1. 2³ × 2² = (2 × 2 × 2) × (2 × 2) = 2\(^{3 + 2}\) = 2⁵

2. 3⁴ × 3² = (3 × 3 × 3 × 3) × (3 × 3) = 3\(^{4 + 2}\) = 3⁶

3. (-3)³ × (-3)⁴ = [(-3) × (-3) × (-3)] × [(-3) × (-3) × (-3) × (-3)]

                        = (-3)\(^{3 + 4}\) 

                        = (-3)⁷

4. m⁵ × m³ = (m × m × m × m × m) × (m × m × m)

                  = m\(^{5 + 3}\) 

                  = m⁸

From the above examples, we can generalize that during multiplication when the bases are same then the exponents are added. 

aᵐ × aⁿ = a\(^{m + n}\)

In other words, if ‘a’ is a non-zero integer or a non-zero rational number and m and n are positive integers, then

aᵐ × aⁿ = a\(^{m + n}\)

Similarly, (\(\frac{a}{b}\))ᵐ × (\(\frac{a}{b}\))ⁿ = (\(\frac{a}{b}\))\(^{m + n}\)

\[(\frac{a}{b})^{m} \times (\frac{a}{b})^{n} = (\frac{a}{b})^{m + n}\]

Note: 

(i) Exponents can be added only when the bases are same. 

(ii) Exponents cannot be added if the bases are not same like

m⁵ × n⁷, 2³ × 3⁴

For example:

1. 5³ ×5⁶

= (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5 × 5)

= 5\(^{3 + 6}\), [here the exponents are added] 

= 5⁹

2. (-7)\(^{10}\) × (-7)¹²

= [(-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7)] × [( -7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7) × (-7)].

= (-7)\(^{10 + 12}\), [Exponents are added] 

= (-7)²²

3. \((\frac{1}{2})^{4}\) × \((\frac{1}{2})^{3}\)

=[(\(\frac{1}{2}\)) × (\(\frac{1}{2}\)) × (\(\frac{1}{2}\)) × (\(\frac{1}{2}\))] × [(\(\frac{1}{2}\)) × (\(\frac{1}{2}\)) × (\(\frac{1}{2}\))] 

=(\(\frac{1}{2}\))\(^{4 + 3}\)

=(\(\frac{1}{2}\))⁷

4. 3² × 3⁵

= 3\(^{2 + 5}\)

= 3⁷

5. (-2)⁷ × (-2)³

= (-2)\(^{7 + 3}\)

= (-2)\(^{10}\)

6. (\(\frac{4}{9}\))³ × (\(\frac{4}{9}\))²

= (\(\frac{4}{9}\))\(^{3 + 2}\)

= (\(\frac{4}{9}\))⁵

We observe that the two numbers with the same base are

multiplied; the product is obtained by adding the exponent.

2. Dividing Powers with the same Base

For example: 

3⁵ ÷ 3¹, 2² ÷ 2¹, 5(²) ÷ 5³

In division if the bases are same then we need to subtract the exponents. 

Consider the following: 

2⁷ ÷ 2⁴ = \(\frac{2^{7}}{2^{4}}\)

            = \(\frac{2 × 2 × 2 × 2 × 2 × 2 × 2}{2 × 2 × 2 × 2}\)

            = 2\(^{7 - 4}\)

            = 2³

5⁶ ÷ 5² = \(\frac{5^{6}}{5^{2}}\)

            = = \(\frac{5 × 5 × 5 × 5 × 5 × 5}{5 × 5}\)

            = 5\(^{6 - 2}\) 

            = 5⁴

10⁵ ÷ 10³ = \(\frac{10^{5}}{10^{3}}\)

                = \(\frac{10 × 10 × 10 × 10 × 10}{10 × 10 × 10}\)

                = 10\(^{5 - 3}\)

                = 10²

7⁴ ÷ 7⁵ = \(\frac{7^{4}}{7^{5}}\)

            = \(\frac{7 × 7 × 7 × 7}{7 × 7 × 7 × 7 × 7}\)

            = 7\(^{4 - 5}\) 

            = 7\(^{-1}\)

Let a be a non zero number, then

a⁵ ÷ a³ = \(\frac{a^{5}}{a^{3}}\)

            = \(\frac{a × a × a × a × a}{a × a × a}\)

            = a\(^{5 - 3}\) 

            = a²

again, a³ ÷ a⁵ = \(\frac{a^{3}}{a^{5}}\)

                     = \(\frac{a × a × a}{a × a × a × a × a}\)

                     = a\(^{-(5 - 3)}\)

                     = a\(^{-2}\)

Thus, in general, for any non-zero integer a, 

aᵐ ÷ aⁿ = \(\frac{a^{m}}{a^{n}}\) = a\(^{m - n}\)

Note 1: 

Where m and n are whole numbers and m > n; 

aᵐ ÷ aⁿ = \(\frac{a^{m}}{a^{n}}\) = a\(^{-(n - m)}\)

Note 2: 

Where m and n are whole numbers and m < n; 

We can generalize that if ‘a’ is a non-zero integer or a non-zero rational number and m and n are positive integers, such that m > n, then 

aᵐ ÷ aⁿ = a\(^{m - n}\) if m < n, then aᵐ ÷ aⁿ = \(\frac{1}{a^{n - m}}\)

Similarly, \((\frac{a}{b})^{m}\) ÷ \((\frac{a}{b})^{n}\) = \(\frac{a}{b}\) \(^{m - n}\)

For example:

1. 7\(^{10}\) ÷ 7⁸ = \(\frac{7^{10}}{7^{8}}\)

                             = \(\frac{7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7 × 7}{7 × 7 × 7 × 7 × 7 × 7 × 7 × 7}\)

                             = 7\(^{10 - 8}\), [here exponents are subtracted] 

                             = 7²


2. p⁶ ÷ p¹ = \(\frac{p^{6}}{p^{1}}\)

               = \(\frac{p × p × p × p × p × p}{p}\)

               = p\(^{6 - 1}\), [here exponents are subtracted] 

               = p⁵


3. 4⁴ ÷ 4² = \(\frac{4^{4}}{4^{2}}\)

                = \(\frac{4 × 4 × 4 × 4}{4 × 4}\)

                = 4\(^{4 - 2}\), [here exponents are subtracted] 

                = 4²


4. 10² ÷ 10⁴ = \(\frac{10^{2}}{10^{4}}\)

                   = \(\frac{10 × 10}{10 × 10 × 10 × 10}\)

                   = 10\(^{-(4 - 2)}\), [See note (2)] 

                   = 10\(^{-2}\)

5. 5³ ÷ 5¹

= 5\(^{3 - 1}\)

= 5²

6. \(\frac{(3)^{5}}{(3)^{2}}\)

= 3\(^{5 - 2}\)

= 3³


7. \(\frac{(-5)^{9}}{(-5)^{6}}\)

= (-5)\(^{9 - 6}\)

= (-5)³


8. (\(\frac{7}{2}\))⁸ ÷ (\(\frac{7}{2}\))⁵

= (\(\frac{7}{2}\))\(^{8 - 5}\)

= (\(\frac{7}{2}\))³

3. Power of a Power

For example: (2³)², (5²)⁶, (3² )\(^{-3}\)

In power of a power you need multiply the powers.

Consider the following

(i) (2³)⁴

Now, (2³)⁴ means 2³ is multiplied four times

i.e. (2³)⁴ = 2³ × 2³ × 2³ × 2³

=2\(^{3 + 3 + 3 + 3}\)

=2¹²

Note: by law (l), since aᵐ × aⁿ = a\(^{m + n}\).

(ii) (2³)²

Similarly, now (2³)² means 2³ is multiplied two times

i.e. (2³)² = 2³ × 2³

= 2\(^{3 + 3}\), [since aᵐ × aⁿ = a\(^{m + n}\)] 

= 2⁶

Note: Here, we see that 6 is the product of 3 and 2 i.e,

                         (2³)² = 2\(^{3 × 2}\)= 2⁶

(iii) (4\(^{- 2}\))³

Similarly, now (4\(^{-2}\))³ means 4\(^{-2}\)

 is multiplied three times

i.e. (4\(^{-2}\))³ =4\(^{-2}\) × 4\(^{-2}\) × 4\(^{-2}\)

= 4\(^{-2 + (-2) + (-2)}\)

= 4\(^{-2 - 2 - 2}\)

= 4\(^{-6}\)

Note: Here, we see that -6 is the product of -2 and 3 i.e,

                (4\(^{-2}\))³ = 4\(^{-2 × 3}\) = 4\(^{-6}\)

For example:

1.(3²)⁴ = 3\(^{2 × 4}\) = 3⁸

2. (5³)⁶ = 5\(^{3 × 6}\) = 5¹⁸

3. (4³)⁸ = 4\(^{3 × 8}\) = 4²⁴

4. (aᵐ)⁴ = a\(^{m × 4}\) = a⁴ᵐ

5. (2³)⁶ = 2\(^{3 × 6}\) = 2¹⁸

6. (xᵐ)\(^{-n}\) = x\(^{m × -(n)}\) = x\(^{-mn}\)

7. (5²)⁷ = 5\(^{2 × 7}\) = 5¹⁴

8. [(-3)⁴]² = (-3)\(^{4 × 2}\) = (-3)⁸

In general, for any non-integer a, (aᵐ)ⁿ= a\(^{m × n}\) = a\(^{mn}\)

Thus where m and n are whole numbers. 

If ‘a’ is a non-zero rational number and m and n are positive integers, then {(\(\frac{a}{b}\))ᵐ}ⁿ = (\(\frac{a}{b}\))\(^{mn}\)

For example:

[(\(\frac{-2}{5}\))³]²

= (\(\frac{-2}{5}\))\(^{3 × 2}\)

= (\(\frac{-2}{5}\))⁶

4. Multiplying Powers with the same Exponents

For example: 3² × 2², 5³ × 7³

We consider the product of 4² and 3², which have different bases, but the same exponents. 

(i) 4² × 3² [here the powers are same and the bases are different] 

= (4 × 4) × (3 × 3) 

= (4 × 3) × (4 × 3) 

= 12 × 12

= 12²

Here, we observe that in 12², the base is the product of bases 4 and 3. 

We consider, 

(ii) 4³ × 2³

= (4 × 4 × 4) × (2 × 2 × 2)

= (4 × 2)× ( 4 × 2) × (4 × 2)

= 8 × 8 × 8

= 8³

(iii) We also have, 2³ × a³

= (2 × 2 × 2) × (a × a × a)

= (2 × a) × (2 × a) × (2 × a)

= (2 × a)³

= (2a)³ [Here 2 × a = 2a]

(iv) Similarly, we have, a³ × b³

= (a × a × a) × (b × b × b)

= (a × b) × (a × b) × (a × b)

= (a × b)³

= (ab)³ [Here a × b = ab]

Note: In general, for any non-zero integer a, b.

aᵐ × bᵐ

= (a × b)ᵐ

= (ab)ᵐ [Here a × b = ab]

aᵐ × bᵐ = (ab)ᵐ

Note: Where m is any whole number.

(-a)³ × (-b)³

= [(-a) × (-a) × (-a)] × [(-b) × (-b) × (-b)]

= [(-a) × (-b)] × [(-a) × (-b)] × [(-a) × (-b)]

= [(-a) × (-b)]³

= (ab)³, [Here a × b = ab and two negative become positive, (-) × (-) = +]

5. Negative Exponents

If the exponent is negative we need to change it into positive exponent by writing the same in the denominator and 1 in the numerator.

If ‘a’ is a non-zero integer or a non-zero rational number and m is a positive integers, then a\(^{-m}\) is the reciprocal of aᵐ, i.e., 

a\(^{-m}\) = \(\frac{1}{a^{m}}\), if we take ‘a’ as \(\frac{p}{q}\) then (\(\frac{p}{q}\))\(^{-m}\) = \(\frac{1}{(\frac{p}{q})^{m}}\) = (\(\frac{q}{p}\))ᵐ

again, \(\frac{1}{a^{-m}}\) = aᵐ

Similarly, (\(\frac{a}{b}\))\(^{-n}\) = (\(\frac{b}{a}\))ⁿ, where n is a positive integer

Consider the following

2\(^{-1}\) = \(\frac{1}{2}\)

2\(^{-2}\) = \(\frac{1}{2^{2}}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{4}\)

2\(^{-3}\) = \(\frac{1}{2^{3}}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{8}\)

2\(^{-4}\) = \(\frac{1}{2^{4}}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\)  = \(\frac{1}{16}\)

2\(^{-5}\) = \(\frac{1}{2^{5}}\) = \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{1}{32}\)

[So in negative exponent we need to write 1 in the numerator and in the denominator 2 multiplied to itself five times as 2\(^{-5}\). In other words negative exponent is the reciprocal of positive exponent] 

For example:

1. 10\(^{-3}\)

= \(\frac{1}{10^{3}}\), [here we can see that 1 is in the numerator and in the denominator 10³ as we know that negative exponent is the reciprocal] 

= \(\frac{1}{10}\) × \(\frac{1}{10}\) × \(\frac{1}{10}\), [Here 10 is multiplied to itself 3 times] 

= \(\frac{1}{1000}\)

2. (-2)\(^{-4}\)

= \(\frac{1}{(-2)^{4}}\) [Here we can see that 1 is in the numerator and in the denominator (-2)⁴] 

= (- \(\frac{1}{2}\)) × (- \(\frac{1}{2}\)) × (- \(\frac{1}{2}\)) × (- \(\frac{1}{2}\)) 

= \(\frac{1}{16}\)

3. 2\(^{-5}\)

= \(\frac{1}{2^{5}}\)

= \(\frac{1}{2}\) × \(\frac{1}{2}\)

= \(\frac{1}{4}\)

4. \(\frac{1}{3^{-4}}\)

= 3⁴

= 3 × 3 × 3 × 3

= 81


5. (-7)\(^{-3}\)

= \(\frac{1}{(-7)^{3}}\)

6. (\(\frac{3}{5}\))\(^{-3}\)

= (\(\frac{5}{3}\))³

7. (-\(\frac{7}{2}\))\(^{-2}\)

= (-\(\frac{2}{7}\))²

6. Power with Exponent Zero

If the exponent is 0 then you get the result 1 whatever the base is. 

For example: 8\(^{0}\), (\(\frac{a}{b}\))\(^{0}\), m\(^{0}\)…....

If ‘a’ is a non-zero integer or a non-zero rational number then, 

a\(^{0}\) = 1

Similarly, (\(\frac{a}{b}\))\(^{0}\) = 1

Consider the following

a\(^{0}\) = 1 [anything to the power 0 is 1] 

(\(\frac{a}{b}\))\(^{0}\) = 1

(\(\frac{-2}{3}\))\(^{0}\) = 1

(-3)\(^{0}\) = 1

For example:

1. (\(\frac{2}{3}\))³ × (\(\frac{2}{3}\))\(^{-3}\)

= (\(\frac{2}{3}\))\(^{3 + (-3)}\), [Here we know that aᵐ × aⁿ = a\(^{m + n}\)] 

= (\(\frac{2}{3}\))\(^{3 - 3}\)

= (\(\frac{2}{3}\))\(^{0}\)

= 1

2. 2⁵ ÷ 2⁵

= \(\frac{2^{5}}{2^{5}}\)

= \(\frac{2 × 2 × 2 × 2 × 2}{2 × 2 × 2 × 2 × 2}\)

= 2\(^{5 - 5}\), [Here by the law aᵐ ÷ aⁿ =a\(^{m - n}\)] 

= 2

= 1

3. 4\(^{0}\) × 3\(^{0}\)

= 1 × 1, [Here as we know anything to the power 0 is 1]

= 1

4. aᵐ × a\(^{-m}\)

= a\(^{m - m}\)

= a\(^{0}\)

= 1

5. 5\(^{0}\) = 1

6. (\(\frac{-4}{9}\))\(^{0}\) = 1

7. (-41)\(^{0}\) = 1

8. (\(\frac{3}{7}\))\(^{0}\) = 1

7. Fractional Exponent

In fractional exponent we observe that the exponent is in fraction form.

a\(^{\frac{1}{n}}\), [Here a is called the base and \(\frac{1}{n}\) is called the exponent or power]

= \(\sqrt[n]{a}\), [nth root of a] 

\[a^{\frac{1}{n}} = \sqrt[n]{a}\]

Consider the following:

2\(^{\frac{1}{1}}\) = 2 (it will remain 2). 

2\(^{\frac{1}{2}}\) = √2 (square root of 2).

2\(^{\frac{1}{3}}\) = ∛2 (cube root of 2).

2\(^{\frac{1}{4}}\) = ∜2 (fourth root of 2).

2\(^{\frac{1}{5}}\) = \(\sqrt[5]{2}\) (fifth root of 2). 

For example:

1. 2\(^{\frac{1}{2}}\) = √2 (square root of 2). 

2. 3\(^{\frac{1}{2}}\) = √3 [square root of 3] 

3. 5\(^{\frac{1}{3}}\) = ∛5 [cube root of 5]

4. 10\(^{\frac{1}{3}}\) = ∛10 [cube root of 10]

5. 21\(^{\frac{1}{7}}\) = \(\sqrt[7]{21}\) [Seventh root of 21]

● Exponents

Exponents

Laws of Exponents

Rational Exponent

Integral Exponents of a Rational Numbers

Solved Examples on Exponents

Practice Test on Exponents

● Exponents - Worksheets

Worksheet on Exponents

8th Grade Math Practice

From Laws of Exponents to HOME PAGE