MULTIPLYING EXPONENTS -
MULTIPLYING POWERS OF
NUMERICALS WITH EXAMPLES, EXERCISE

Please study
Laws of Exponents before Multiplying Exponents,
if you have not already done so.

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 2⁶

Solution:
(i)(3⁄4)⁴ x (2⁄3)⁵ x (3⁄8)²

(3⁄4)⁴ x (2⁄3)⁵ x (3⁄8)²
= (3⁴⁄4⁴) x (2⁵⁄3⁵) x (3²⁄8²)
{since (a⁄b)^m = a^m⁄b^m} (See Law 7 above)
= (3⁴ x 4^-4) x (2⁵ x 3^-5) x (3² x 8^-2)
(since dividing with a^m is same as multiplying with a^-m ) (See Law 3 above)
= (3⁴ x 3^-5 x 3²) x (4^-4 x 2⁵ 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 { (2²)^-4 x 2⁵ x (2³)^-2}
(since a^m x a^-n x a^p = a^{m + (-n) + p}
and 4 = 2² and 8 = 2³)

= ( 3¹) x { 2^{2 x -4} x 2⁵ x 2^{3 x -2}}
{since (a^m)^-n = a^{m x -n}}
= 3 x { 2^-8x 2⁵ x 2^-6}
= 3 x 2^{(-8) + 5 + (-6)} = 3 x 2^-9
= 3⁄2⁹ = 3⁄512. Ans.

(ii) { (1⁄4)² - (1⁄4)³} x 2⁶

{ (1⁄4)² - (1⁄4)³} x 2⁶
= { (1⁄2²)² - (1⁄2²)³} x 2⁶
= { (2^-2)² - (2^-2)³} x 2⁶
= { 2^{-2 x 2} - 2^{-2 x 3}} x 2⁶
= { 2^-4 - 2^-6} x 2⁶
= {2^-4x 2⁶- 2^-6x 2⁶}
= {2^{-4 + 6} - 2^{-6 + 6}}
= {2² - 2⁰}
= 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 bⁿ.
∴ x = 64 x 4³
= 64 x (4 x 4 x 4) = 64 x 64 = 4096. Ans.

Solved Example 3 : Multiplying Exponents

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 bⁿ.
∴ 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.

Solved Example 4 : Multiplying Exponents

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.

For More Solved Examples on application
of Laws of Exponents, go to
Addition of Exponents.

Exercise : Multiplying Exponents

1. Simplify (1⁄2)² x (-2⁄3)³ x (-4⁄5)⁴
2. Find the value of (4⁰ - 3⁰) x 6⁰
3. Fill in the blanks (i)5⁴ x 5⁷ = 5^..... (ii) (5⁄2)⁶ x (2⁄5)⁶ = .........
4. 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.

Answers to Exercise : Multiplying Exponents

1. -512⁄16875
2. 0
3. (i) 11 (ii) 1
4. -7⁄10