The laws of exponents are explained here along with their examples.
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.
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
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
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, (-) × (-) = +]
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
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
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
Integral Exponents of a Rational Numbers
● Exponents - Worksheets
8th Grade Math Practice
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