FACTORING EXPONENTS -
EXPRESSING NUMBERS AS A
PRODUCT OF PRIME FACTORS
IN EXPONENTIAL NOTATION

Please study
Basics of Exponents before Factoring Exponents
if you have not already done so.

There, we introduced the
special notation of writing a
(literal) number being multiplied
more than once, with examples.

We explained the terms :
Base, Exponent or Index,
Exponential form, Expanded form.

Expressing the given numbers
as a product of prime factors
in the exponential notation is
an essential knowledge in both
Exponents and Prime Factorization.

Here, in Factoring Exponents,
we deal with that with solved
Examples and exercise for practice.

Solved Example 1 : Factoring Exponents

Express the following numbers
as a product of prime factors
in the exponential notation.
(i) 96 (ii) 144 (iii) 2250 (iv) 504

Solution : Factoring Exponents
(i) 96
We know, if the last digit
is divisible by 2, the number
is divisible by 2.
96⁄2 = 48; 48⁄2 = 24; 24⁄2 = 12; 12⁄2 = 6; 6⁄2 = 3;
∴ 96 = 2 x48 = 2 x 2 x 24 = 2 x 2 x 2 x 12 = 2 x 2 x 2 x 2 x 6
= 2 x 2 x 2 x 2 x 2 x 3 = 2⁵ x 3. Ans.

(ii) 144
proceeding as in (i) above, we get
144 = 2 x 72 = 2 x 2 x 36 = 2 x 2 x 2 x 18 = 2 x 2 x 2 x 2 x 9
= 2 x 2 x 2 x 2 x 3 x 3 = 2⁴ x 3². Ans.

(iii) 2250
2250 = 2 x 1125.
We know, if the last digit is 5 or 0, the number is divisible by 5.
So, by dividing 1125 with 5, we get 225. ie 1125 = 5 x 225.
∴ 2250 = 2 x 1125 = 2 x 5 x 225
= 2 x 5 x 5 x 45 (since 225⁄5 = 45) = 2 x 5 x 5 x 5 x 9 (since 45⁄5 = 9)
= 2 x 5 x 5 x 5 x 3 x 3 = 2 x 5³ x 3². Ans.

(iv) 504
504 = 2 x 252 = 2 x 2 x 126 = 2 x 2 x 2 x 63.
We know, if the sum of the digits is divisible by 3,
the number is divisible by 3.
So 63 is divisible by 3 (since 6 + 3 = 9 is divisible by 3).
63⁄3 = 21
∴504 = 2 x 2 x 2 x 3 x 21 = 2 x 2 x 2 x 3 x 3 x 7
= 2³ x 3² x 7¹. Ans.

Solved Example 2 : Factoring Exponents

Express in power notation:
(i) 1⁄243 (ii) 27⁄125 (iii) -64⁄25 (iv) 125⁄343

Solution : Factoring Exponents
(i) 1⁄243
243 = 3 x 81 = 3 x 3 x 27 = 3 x 3 x 3 x 9 = 3 x 3 x 3 x 3 x 3
= 3⁵
1⁄243 = 1⁄3⁵
We know, 1⁄aⁿ = a^-n (see Law 3 above)
∴1⁄243 = 1⁄3⁵ = 3^-5. Ans.

(ii) 27⁄125
27 = 3 x 9 = 3 x 3 x 3 = 3³
125 = 5 x 25 = 5 x 5 x 5 = 5³
27⁄125 = 3³⁄5³
We know,(a⁄b)^m = a^m⁄b^m (See Law 7 above)
∴27⁄125 = 3³⁄5³ = (3⁄5)³. Ans.

(iii) -64⁄25
25 = 5 x 5 = 5²
Since denominator is of power 2,
Let us put numerator also as power 2.
-64 = -8 x 8 = -8²
Note that -8² is not same as (-8)².
-64⁄25 = -8²⁄5²
We know,(a⁄b)^m = a^m⁄b^m (See Law 7 above)
∴-64⁄25 = -8²⁄5² = -(8⁄5)². Ans.
Note that '-' is out side the bracket.

(iv) 125⁄343
125 = 5 x 25 = 5 x 5 x 5 = 5³
What about 343?
Did you try the problem 1 (vi) of exercise in







The Exponents ?
343 is not divisible by the primes 2, 3, 5.
What about 7?

To find out if a number is divisible by seven,
take the last digit, double it, and subtract it from the rest of the number.
Example: If you had 203, you would double the last digit to get six,
and subtract that from 20 to get 14.
If you get an answer divisible by 7 (including zero),
then the original number is divisible by seven.
If you don't know the new number's divisibility,
you can apply the rule again.

Here, in Factoring Exponents,
we made use of the knowledge of
Divisibilty Rules
and
Prime Factorization.

For study of more Examples
on exponents, go to
Multiplying Exponents

Exercise : Factoring Exponents

1. Express 8⁄125 and -8⁄125 as powers of a rational number.
2. Express 8⁄27 in power notation.

For Answers, see at the bottom of the page.

For more problems for practice on exponents, go to
Multiplying Exponents

Answers to Exercise : Factoring Exponents

1. (2⁄5 )³ and (-2⁄5 )³
2. (2⁄3 )³