First Law
If a is any non-zero rational number and m, n are natural numbers, then
\( { a }^{ m } \) x \( { a }^{ n } \) = \( { a }^{ m + n } \)
Generalised form of above law:
If a is a non-zero rational number and m, n, p are natural numbers, then,
\( { a }^{ m } \) x \( { a }^{ n } \) x \( { a }^{ p } \) = \( { a }^{ m + n + p } \)
Illustration : Simplify and write the answer of each of the following in exponential form:
(i) \( { 4 }^{ 2 } \) x \( { 4 }^{ 3 } \) (ii) \( { 2 }^{ 2 } \) x \( { 2 }^{ 3 } \) x \( { 2 }^{ 4 } \)
(iii) \( { 6 }^{ x } \) x \( { 6 }^{ 3 } \) (iv) \( { (\frac { 3 }{ 2 } ) }^{ 3 } \) x \( { (\frac { 3 }{ 2 } ) }^{ 6 } \)
Solution. Using first law of exponents, We have
(i) \( { 4 }^{ 2 } \) x \( { 4 }^{ 3 } \) = \( { 4 }^{ 2 + 3 } \) = \( { 4 }^{ 5 } \)
(ii) \( { 2 }^{ 2 } \) x \( { 2 }^{ 3 } \) x \( { 2 }^{ 4 } \) = \( { 2 }^{ 2 + 3 + 4 } \) = \( { 2 }^{ 9 } \)
(iii) \( { 6 }^{ x } \) x \( { 6 }^{ 3 } \) = \( { 6 }^{ x + 3 } \)
(iv) \( { (\frac { 3 }{ 2 } ) }^{ 3 } \) x \( { (\frac { 3 }{ 2 } ) }^{ 6 } \)= \( { (\frac { 3 }{ 2 } ) }^{ 6 + 3 } \) = \( { (\frac { 3 }{ 2 } ) }^{ 9 } \)
Second Law
If a is any non-zero rational number and m and n are natural numbers such that m > n, then
\( { a }^{ m } \) \( \div \) \( { a }^{ n } \) = \( { a }^{ m – n } \) or \( \frac { { a }^{ m } }{ { a }^{ n } } \) = \( { a }^{ m – n } \)
Illustration : Simplify and write each of the following in exponential form:
(i) \( { 8 }^{ 6 } \) \( \div \) \( { 8 }^{ 3 } \) (ii) \( { (-5) }^{ 10 } \) ÷ \( { (-5) }^{ 4 } \)
(iii) \( { (\frac { -3 }{ 5 } ) }^{ 6 } \) ÷ \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)
Solution. Using second law of exponents, We have
(i) \( { 8 }^{ 6 } \) \( \div \) \( { 8 }^{ 3 } \)
= \( \frac { { 8 }^{ 6 } }{ { 8 }^{ 3 } } \)
= \( { 8 }^{ 6 – 3 } \)
= \( { 8 }^{ 3 } \)
(ii) \( { (-5) }^{ 10 } \) ÷ \( { (-5 )}^{ 4 } \)
= \( \frac { { (-5) }^{ 10 } }{ {(-5) }^{ 4 } } \)
= \( { (-5) }^{ 10 – 4 } \)
= \( { (-5) }^{ 6 } \)
(iii) \( { (\frac { -3 }{ 5 } ) }^{ 6 } \) ÷ \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)
= \( { \frac { { (\frac { -3 }{ 5 } })^{ 6 } }{ ({ \frac { -3 }{ 5 } ) }^{ 3 } } } \)
= \( { (\frac { -3 }{ 5 } ) }^{ 6 – 3 } \)
= \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)
Third Law
If a is any rational number different from zero and m, n are natural numbers, then
\( { { (a }^{ m }) }^{ n }\quad =\quad { a }^{ m\quad \times \quad n }\quad =\quad { { (a }^{ n }) }^{ m } \)Illustration: Simplify and write each of the following in exponential form:
(i) \( { { (3 }^{ 2 }) }^{ 4 }\quad \) (ii) \( { { ((-2) }^{ 4 }) }^{ 2 }\quad \)
(iii) \( { { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 } \) (iv) \( { ({ 4 }^{ 2 }) }^{ 3 }\quad \) x \( ({ { 4}^{ 4 }) }^{ 2 }\quad \)
Solution. Using third law of exponents, We have
(i) \( { { (3 }^{ 2 }) }^{ 4 }\quad \)
= \( { 3 }^{ 2 \times 4 } \)
= \( { 3 }^{ 8 } \)
(ii) \( { { ((-2) }^{ 4 }) }^{ 2 }\quad \)
= \( { (-2) }^{ 4 \times 2 } \)
= \( { (-2) }^{ 8 } \)
(iii) \( { { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 } \)
= \( { (\frac { 3 }{ 5 } ) }^{ 3 \times 4 } \)
= \( { (\frac { 3 }{ 5 } ) }^{ 12 } \)
(iv) \( { ({ 4 }^{ 2 }) }^{ 3 }\quad \)x \( { ({ 4}^{ 4 }) }^{ 2 }\quad \)
= \( ({ 4 }^{ 2 \times 3}) \) x \( ({ 4 }^{ 4 \times 2 }) \)
= \( { 4 }^{ 6 } \) x \( { 4 }^{ 8 } \)
= \( { 4 }^{ 6 + 8 } \)
= \( { 4 }^{ 14 } \)
Fourth Law
If a, b are non-zero rational numbers and n is a natural number, then
\( { a }^{ n } \) x \( { b }^{ n } \) = \( { (ab) }^{ n } \)
Generalised form of above law:
If a, b, c are non-zero rational numbers and n is a natural number, then
\( { a }^{ n } \) x \( { b }^{ n } \) x \( { c }^{ n } \) = \( {(abc)}^{ n } \)
Illustration: Express each of the following products of powers as the exponent of a rational number:
(i) \( { 2 }^{ 4 } \) x \( { 5 }^{ 4 } \) (ii) \( {(-3) }^{ 3 } \) x \( { (-2) }^{ 3 } \)
(iii) \( { 3 }^{ 2 } \) x \( { x }^{ 2 } \) x \( { y }^{ 2 } \) (iv) \( { (\frac { 3 }{ 2 } ) }^{ 2 } \) x \( { (\frac { 2 }{ 5 } ) }^{ 2 } \)
Solution. We have,
(i) \( { 2 }^{ 4 } \) x \( { 5 }^{ 4 } \)
= \( { (2 \times 5 )}^{ 4 } \)
= \( { 10 }^{ 4 } \)
(ii) \( {(-3) }^{ 3 } \) x \( { (-2) }^{ 3 } \)
= \( { ((-3) \times (-2) )}^{ 3 } \)
= \( { 6 }^{ 3 } \)
(iii) \( { 3 }^{ 2 } \) x \( { x }^{ 2 } \) x \( { y }^{ 2 } \)
= \( {( 3 \times x \times y )}^{ 2 } \)
= \( { (3xy) }^{ 2 } \)
(iv) \( { (\frac { 3 }{ 2 } ) }^{ 2 } \) x \( { (\frac { 2 }{ 5 } ) }^{ 2 } \)
= \( { (\frac { 3 }{ 2 } }\times \frac { 2 }{ 5 } )^{ 2 } \)
= \( { (\frac { 3 }{ 5 } ) }^{ 2 } \)
Fifth Law
If a and b are non-zero rational numbers and n is a natural number, then
\( \frac { { a }^{ n } }{ { b }^{ n } } \) = \( { (\frac { a }{ b } ) }^{ n } \)
Illustration: Write each of the following in the form \(\frac{p}{q}\)
(i) \( { (\frac { 2 }{ 5 } ) }^{ 3 } \) (ii) \( { (\frac { -3 }{ 2 } ) }^{ 4 } \)
Solution. We have,
(i) \( { (\frac { 2 }{ 5 } ) }^{ 3 } \)
= \( \frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } } \)
= \(\frac{2 \times 2 \times 2}{5 \times 5 \times 5 }\)
= \(\frac{8}{125}\)
(ii) \( { (\frac { -3 }{ 2 } ) }^{ 4 } \)
= \( \frac { { (-3) }^{ 4 } }{ { 2 }^{ 4 } } \)
= \(\frac{(-3) \times (-3) \times (-3) \times (-3)}{2 \times 2 \times 2 \times 2}\)
= \(\frac{81}{16}\)
Illustrative Examples
Example 1: Use the laws of exponents to simplify the following :
(i) \( { [{ (2^{ 3 }) }^{ 4 }] }^{ 5 } \) (ii) \( [{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 } \)
(iii) \( { 81 }^{ -1 } \) x \( { 3 }^{ 5 } \) (iv) \( { (\frac { 2 }{ 3 } ) }^{ 0 } \) + \( { (\frac { 2 }{ 3 } ) }^{ -2 } \)
Solution. We have,
(i) \( { [{ (2^{ 3 }) }^{ 4 }] }^{ 5 } \)
= \( { [{ 2^{ 3 \times 4 }}] }^{ 5 } \)
= \( { [{ 2^{ 12 }}] }^{ 5 } \)
= \( { [{ 2^{ 12 \times 5}}] } \)
= \( { 2 }^{ 60 } \)
(ii) \( [{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 } \)
= \( { [ 3^{ 6 – 4 }]^3} \)
= \( { [{ 3^{ 2 }}] }^{ 3 } \)
= \( { 3 }^{ 2 \times 3 } \)
= \( { 3 }^{ 6 } \)
(iii) \( { 81 }^{ -1 } \) x \( { 3 }^{ 5 } \)
= \( { { (3 }^{ 4 }) }^{ -1 }\quad \) x \( { 3 }^{ 5 } \)
= \( { 3 }^{ -4 + 5 } \)
= \( { 3 }^{ 1 } \)
= 3
(iv) \( { (\frac { 2 }{ 3 } ) }^{ 0 } \) + \( { (\frac { 2 }{ 3 } ) }^{ -2 } \)
= 1 + \( { \frac { 1 }{ ({ \frac { 2 }{ 3 } ) }^{ 2 } } } \)
= \( { \frac { 1 }{ \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } } } } \)
= \( { \frac { 1 }{ \frac { 4 }{ 9 } } } \)
= 1 + \(\frac{9}{4}\)
= \(\frac{13}{4}\)
Example 2: Simplify and write the answer in the exponential form:
(i) \( { ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad } \) (ii) \( { (-4) }^{ 3 } \) x \( { (5) }^{ -3 } \) x \( { (-5) }^{ -3 } \)
Solution.
(i) We have,
\( { ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad } \)
= \( { (\frac { { 2 }^{ 5 } }{ { 2 }^{ 8 } } ) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { { (2 }^{ 5 – 8 }) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { { (2 }^{ -3 }) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -3 \times 5 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -15 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -15-5 } \) = \( { (2) }^{ -20 } \)
(ii) We have,
\( { (-4) }^{ 3 } \) x \( { (5) }^{ -3 } \) x \( { (-5) }^{ -3 } \)
= \( { [-4 \times 5 \times (-5)] }^{ -3 } \)
= \( { (100) }^{ -3 } \)
= \( { { (10 }^{ 2 }) }^{ -3 }\quad \)
= \( { (10) }^{ 2 \times -3 } \) = \( { (10) }^{ -6 } \)
Example 3: Simplify
(i) \( { ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }} \) (ii) \( { ({ { 5 }^{ 3 } \times { 3 }^{ -1 }) }^{ -1 }} \) ÷ \( { 6 }^{ -1 } \)
Solution.
(i) \( { ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }} \)
= \( { (\frac { 1 }{ 4 } \times \frac { 3 }{ 1 } ) }^{ -2 } \)
= \( { (\frac { 3 }{ 4 })^{-2}} \)
= \( { (\frac { 4 }{ 3 })^{2}} \)
= \(\frac{16}{9}\)
(ii) \( { ({ { 5 }^{ -1 } \times { 3 }^{ -1 }) }^{ -1 }} \) ÷ \( { 6 }^{ -1 } \)
= \( { (\frac { 1 }{ 5 } \times \frac { 1 }{ 3 } ) }^{ -1 } \) ÷ \(\frac{1}{6}\)
= \( { (\frac { 1 }{ 15 })^ {-1} } \) ÷ \(\frac{1}{6}\)
= \(\frac{15}{1}\) ÷ \(\frac{1}{6}\)
= \(\frac{15}{1} \times \frac{6}{1}\)
= 90
Example 4: Find the value of m if \( { (\frac { 2 }{ 9 })^ {3} } \) x \( { (\frac { 2 }{ 9 })^ {-6} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)
Solution.
\( { (\frac { 2 }{ 9 })^ {3} } \) x \( { (\frac { 2 }{ 9 })^ {-6} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)
=> \( { (\frac { 2 }{ 9 })^ {3+(-6)} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)
=> \( { (\frac { 2 }{ 9 })^ {-3} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)
In an equation, when bases on both sidess are equal, their powers must also be equal.
Therefore, 2m – 1 = -3 or 2m = -3 + 1
=> 2m = -2
=> m = \(\frac{-2}{2}\) = –1
Example 5: What number should \( { (\frac { 2 }{ 3 })^ {-2} } \) be multiplied so that the product is \( { (\frac { 4 }{ 27 })^ {-1} } \) ?
Solution.
\( x \) x \( { (\frac { 2 }{ 3 })^ {-2} } \) = \( { (\frac { 4 }{ 27 })^ {-1} } \)
=> \( x \) x \( { (\frac { 3 }{ 2 })^ {2} } \) = \(\frac{27}{4}\)
=> \( x \) x \(\frac{9}{4}\) = \(\frac{27}{4}\)
=> \( x \) = \(\frac{27}{4}\) x \(\frac{4}{9}\)
=> \( x \) = 3.
Hence, the required number is 3.
Example 6: Simplify
\( \frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 } \)
Solution. We have,
\( \frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 } \)
= \( \frac { ({ 2 }^{ 2 }\times 3)^{4} \times ({ 3 }^{ 2 }) ^ {3} \times {2}^{2} }{ { {2}^ {3}\times 3 }^{ 3 }\times ({ 2 }^{ 3 })^{2}\times {3}^{3}) } \)
= \( \frac { { ({ 2 }^{ 2 }) }^{ 4 }\times { 3 }^{ 4 }\times { ({ 3 }^{ 2 }) }^{ 3 }\times { 2 }^{ 2 } }{ ({ 2 }^{ 3 }\times { 3 }^{ 3 })\times { { (2 }^{ 3 }) }^{ 2 }\times { 3 }^{ 3 } } \)
= \( \frac { { 2 }^{ 8 }\times { 3 }^{ 4 }\times {3}^{6} \times {2}^{2} }{ { 2 }^{ 3 }\times { 3 }^{ 3 }\times {2}^{6} \times {3}^{3} } \)
= \( \frac { ({ 2 }^{ 8 }\times { 2 }^{ 2 } )\times ({3}^{6} \times {3}^{4}) }{ ({ 2 }^{ 3 }\times { 2 }^{ 6 }\times ({3}^{3} \times {3}^{3}) } \)
= \( \frac { { 2 }^{ 8+2 }\times { 3 }^{ 4+6 } }{ { 2 }^{ 3+6 }\times { 3 }^{ 3+3 } } \)
= \( \frac { { 2 }^{ 10 }\times { 3 }^{ 10 } }{ { 2 }^{ 9 }\times { 3 }^{ 6 } } \)
= \( \frac { { 2 }^{ 10 } }{ { 2 }^{ 9 } } \) x \( \frac { { 3 }^{ 10 } }{ { 3 }^{ 6 } } \)
= \( { 2 }^{ 10-9 } \) x \( { 3 }^{ 10-6 } \)
= \( { 2 }^{ 1 } \) x \( { 3 }^{ 4 } \)
= 2 x 81 = 162
Example 7: Find the values of n in each of the following:
(i) \( { ({ 2 }^{ 2 }) }^{ n } \) = \( { ({ 2 }^{ 3 }) }^{ 4 } \) (ii) \( { 2 }^{ 5n } \) ÷ \( { 2 }^{ n } \) = \( { 2 }^{ 4 } \)
Solution.
(i) We have,
\( { ({ 2 }^{ 2 }) }^{ n } \) = \( { ({ 2 }^{ 3 }) }^{ 4 } \)
=> \( { 2 }^{ 2n } \) = \( { 2 }^{ 3 \times 4 } \)
=> \( { 2 }^{ 2n } \) = \( { 2 }^{ 12 } \)
=> 2n = 12 [On equating the exponents]
=> n = \(\frac{12}{2}\) = 6
(ii) We have,
\( { 2 }^{ 5n } \) ÷ \( { 2 }^{ n } \) = \( { 2 }^{ 4 } \)
=> \( \frac { { 2 }^{ 5n } }{ { 2 }^{ n } } \) = \( { 2 }^{ 4 } \)
=> \( { 2 }^{ 5n-n } \) = \( { 2 }^{ 4 } \) [since, \( \frac { { a }^{ m } }{ { a }^{ n } } \) = \( { a }^{ m-n } \) ]
=> \( { 2 }^{ 4n } \) = \( { 2 }^{ 4 } \)
=> 4n = 4 [On equating the exponents]
=> n = \(\frac{4}{4}\) = 1
Example 8: If \({25}^{n-1}\) + 100 = \({5}^{2n-1}\), find the value of n.
Solution. We have,
\({25}^{n-1}\) + 100 = \({5}^{2n-1}\)
=> \({5}^{2n-1}\) – \({5}^{2n-1}\) = 100
=> \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { { 25 }^{ n } }{ { 25} } \) = \({10}^{2}\)
=> \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { ({ 5 } ^{2})^{ n } }{ { 25} } \) = \( ({2 \times 5}) ^{2}\)
=> \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \(\frac{1}{5}\) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \((\frac{1}{5})\) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \((\frac{5 – 1}{25})\) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \((\frac{4}{25})\) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \( \frac { { 2 }^{ 2 } }{ { 5 }^{ 2 } } \) = \({2}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) x \({2}^{2}\) = \({2}^{2} \times {5}^{2} \times {5}^{2}\)
=> \({5}^{2n}\) = \( \frac { { 2 }^{ 2 } \times {5}^{2} \times {5}^{2} }{ { 2 }^{ 2 } } \)
=> \({5}^{2n}\) = \({2}^{2 – 2}\) x \({5}^{2 + 2}\)
=> \({5}^{2n}\) = \({2}^{0}\) x \({5}^{4}\)
=> \({5}^{2n}\) = \({5}^{4}\)
=> 2n = 4 [On equating the exponents]
=> n = \(\frac{4}{2}\) = 2