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 = 2^{5} 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 = 2^{4} x 3^{2}. 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 5^{3} x 3^{2}. 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 = 2^{3} x 3^{2} x 7^{1}. Ans.

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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 = 3^{5} 1⁄243 = 1⁄3^{5} We know, 1⁄a^{n} = a^{-n} (see Law 3 above) ∴1⁄243 = 1⁄3^{5} = 3^{-5}. Ans.

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

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

(iv) 125⁄343 125 = 5 x 25 = 5 x 5 x 5 = 5^{3} 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.

Apply this divisibility by 7 rule to 343. 34 - 2 x 3 = 34 - 6 = 28 is divisible by 7. So 343 is divisible by 7. 343⁄7 = 49. ∴343 = 7 x 49 = 7 x 7 x 7 = 7^{3} 125⁄343 = 5^{3}⁄7^{3} = (5⁄7)^{3}. Ans.

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