There, we state the 7 Laws of indices and
the two Rules used for solving problems.

We also provide Links to the
explanations and proofs of the 7 laws.

Here, in Multiplying Exponents,
we apply the 7 Laws and the
2 Rules in solving problems.

Solved Example 1 : Multiplying Exponents

Find the value of
(i) (3⁄4)^{4} x (2⁄3)^{5} x (3⁄8)^{2}
(ii) { (1⁄4)^{2} - (1⁄4)^{3}} x 2^{6}

Solution:
(i)(3⁄4)^{4} x (2⁄3)^{5} x (3⁄8)^{2}

(3⁄4)^{4} x (2⁄3)^{5} x (3⁄8)^{2}
= (3^{4}⁄4^{4}) x (2^{5}⁄3^{5}) x (3^{2}⁄8^{2})
{since (a⁄b)^{m} = a^{m}⁄b^{m}} (See Law 7 above)
= (3^{4} x 4^{-4}) x (2^{5} x 3^{-5}) x (3^{2} x 8^{-2})
(since dividing with a^{m} is same as multiplying with a^{-m} ) (See Law 3 above)
= (3^{4} x 3^{-5} x 3^{2}) x (4^{-4} x 2^{5} x 8^{-2})
(obtained by putting all factors of base 3 at one place and factors of base 4, 2, 8 at one place)
= (3^{4 + (-5) + 2}) x { (2^{2})^{-4} x 2^{5} x (2^{3})^{-2}}
(since a^{m} x a^{-n} x a^{p} = a^{m + (-n) + p}
and 4 = 2^{2} and 8 = 2^{3})

= ( 3^{1}) x { 2^{2 x -4} x 2^{5} x 2^{3 x -2}}
{since (a^{m})^{-n} = a^{m x -n}}
= 3 x { 2^{-8}x 2^{5} x 2^{-6}}
= 3 x 2^{(-8) + 5 + (-6)} = 3 x 2^{-9}
= 3⁄2^{9} = 3⁄512. Ans.

(ii) { (1⁄4)^{2} - (1⁄4)^{3}} x 2^{6}

{ (1⁄4)^{2} - (1⁄4)^{3}} x 2^{6}
= { (1⁄2^{2})^{2} - (1⁄2^{2})^{3}} x 2^{6}
= { (2^{-2})^{2} - (2^{-2})^{3}} x 2^{6}
= { 2^{-2 x 2} - 2^{-2 x 3}} x 2^{6}
= { 2^{-4} - 2^{-6}} x 2^{6}
= {2^{-4}x 2^{6}- 2^{-6}x 2^{6}}
= {2^{-4 + 6} - 2^{-6 + 6}}
= {2^{2} - 2^{0}}
= 4 - 1 = 3. Ans.

Solved Example 2 : Multiplying Exponents

By what number should we multiply 4^{-3}
so that the product may be equal to 64?

Solution:
Let x be the number with which we should multiply 4^{-3}
so that the product may be equal to 64.
Then x x 4^{-3} = 64.
⇒ x = 64⁄4^{-3}
We know a⁄b^{-n} = a x b^{n}.
∴ x = 64 x 4^{3}
= 64 x (4 x 4 x 4) = 64 x 64 = 4096. Ans.

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By what number should (-2)^{-3} be multiplied
so that the product may be equal to 10^{-1}?

Solution:
Let x be the number with which (-2)^{-3} be multiplied
so that the product may be equal to 10^{-1}
Then x x (-2)^{-3} = 10^{-1}.
⇒ x = 10^{-1}⁄(-2)^{-3}
We know a⁄b^{-n} = a x b^{n}.
∴ x = 10^{-1} x (-2)^{3}
We know 10^{-1} = 1⁄10 and (-2)^{3} = -8
∴ x = (1⁄10) x -8 = -8⁄10 = -4⁄5. Ans.

By what number should (-25)^{-1} be divided
so that the quotient may be 5^{-1}?

Solution:
Let x be the number with which (-25)^{-1} be divided
so that the quotient may be 5^{-1}
Then (-25)^{-1}⁄x = 5^{-1}.
⇒ (-25)^{-1}⁄5^{-1} = x.
⇒ {1⁄(-25)}⁄{1⁄(5)} = x.
⇒ 1⁄(-25) x 5 = x
⇒ x = -5⁄25 = -1⁄5. Ans.

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Solve the following problems on Multiplying Exponents :

Simplify (1⁄2)^{2} x (-2⁄3)^{3} x (-4⁄5)^{4}

Find the value of (4^{0} - 3^{0}) x 6^{0}

Fill in the blanks (i)5^{4} x 5^{7} = 5^{.....} (ii) (5⁄2)^{6} x (2⁄5)^{6} = .........

Find the number to be multiplied by (-7)^{-1} so as to get (10)^{-1}.

For Answers, see at the bottom of the page.

For more problems for practice on Application of Laws of exponents, go to Addition of Exponents.

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