# 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$$^{\frac{1}{5}}$$ = $$\sqrt{2}$$ (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$$^{\frac{1}{7}}$$ = $$\sqrt{21}$$ [Seventh root of 21]

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