1. Using laws of exponents, simplify and write the answer in exponential form:
(i) \(3^2 × 3^4 × 3^8\)
Solution:-
\(= (3)^{2 + 4 + 8} \ \ \ \ \ \ \ (\because a^m \times a^n = a^{m + n})\)
\(= 3^{14}\)
(ii) \(6^{15} ÷ 6^{10}\)
Solution:-
\(= (6)^{15 – 10} \ \ \ \ \ \ \ (\because a^m ÷ a^n = a^{m – n})\)
\(= 65\)
(iii) \(a^3 × a^2\)
Solution:-
\(= (a)^{3 + 2} \ \ \ \ \ \ \ (\because a^m ÷ a^n = a^{m + n})\)
\(= a^5\)
(iv) \(7^x × 7^2\)
Solution:-
\(= (7)^{x + 2}\)
(v) \((5^2)^3 ÷ 5^3\)
Solution:-
\((5^2)^3 = (5)^{2 × 3}\ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= 5^6\)
\(5^6 ÷ 5^3= (5)^{6 – 3} \ \ \ \ \ \ \ (\because a^m ÷ a^n = a^{m - n})\)
\(= 5^3\)
(vi) \(2^5 × 5^5\)
Solution:-
\(= (2 × 5)^5 \ \ \ \ \ \ \ (\because a^m × b^m = ab^m\)
\(= 10^5\)
(vii) \(a^4 × b^4\)
Solution:-
\(= (a × b)^4 \ \ \ \ \ \ \ (\because (a^m)^n = a^{mn}\)
\(= ab^4\)
(viii) \((3^4)^3\)
Solution:-
\((3^4)^3 = (3)^{4 × 3} \ \ \ \ \ \ \ (\because (a^m)^n = a^{mn}\)
\(= 3^{12}\)
(ix) \((2^{20} ÷ 2^{15}) × 2^3\)
Solution:-
\((2^{20} ÷ 2^{15}) = (2)^{20 – 15}\ \ \ \ \ \ \ (\because a^m ÷ a^n = a^{m – n})\)
\(= 2^5\)
\(2^5 × 2^3 = (2)^{5 + 3} \ \ \ \ \ \ \ (\because a^m \times a^n = a^{m + n})\)
\(= 2^8\)
(x) \(8^t ÷ 8^2\)
Solution:-
\(= (8)^{t – 2} \ \ \ \ \ \ \ (\because a^m ÷ a^n = a^{m – n})\)
2. Simplify and express each of the following in exponential form:
(i) \((2^3 × 3^4 × 4)\over (3 × 32)\)
Solution:-
Factors of \(32 = 2 × 2 × 2 × 2 × 2 = 2^5\)
Factors of \(4 = 2 × 2 = 2^2\)
\((2^3 × 3^4 × 2^2)\over (3 × 2^5)\)
\(= {(2^{3 + 2} × 3^4) \over (3 × 5^2)} \ \ \ \ (∵a^m × a^n = a^{m + n}]\)
\(= {(2^5 × 3^4) \over (3 × 2^5)}\)
\(= 2^{5 – 5} \times 3{4 – 1} \ \ \ \ (∵a^m ÷ a^n = a{m – n}]\)
\(= 2^0 × 3^3\)
\(= 1 × 3^3\)
\(= 3^3\)
(ii) \(((5^2)^3 × 5^4) ÷ 5^7\)
Solution:-
\((5^2)^3 = (5)^{2 × 3} \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= 5^6\)
\(= (5^6 × 5^4) ÷ 5^7\)
\(= (5)^{6 + 4} ÷ 5^7 \ \ \ \ (∵a^m × a^n = a^{m + n}]\)
\(= 5^{10} ÷ 5^7\)
\(= 5^{10 – 7} \ \ \ \ (∵a^m ÷ a^n = a^{m – n}]\)
\(= 5^3\)
(iii) \(25^4 ÷ 5^3\)
Solution:-
\((25)^4 = (5 × 5)^4 = (5^2)^4\)
\((5^2)^4 = (5)^{2 × 4}\ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= 5^8\)
\(= 5^8 ÷ 5^3\)
\(= 5^{8 – 3} \ \ \ \ (∵a^m ÷ a^n = a^{m – n}]\)
\(= 5^5\)
(iv) \((3 × 7^2 × 11^8)\over (21 × 11^3)\)
Solution:-
Factors of \(21 = 7 × 3\)
\(= {(3 × 7^2 × 11^8)\over (7 × 3 × 11^3)}\)
\(= 3^{1-1} × 7^{2-1} × 11^{8 – 3}\)
\(= 3^0 × 7 × 11^5\)
\(= 1 × 7 × 11^5\)
\(= 7 × 11^5\)
(v) \(3^7\over {(3^4 × 3^3)}\)
Solution:-
\(= {3^7\over (3^{4+3})} \ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= {3^7\over 3^7}\)
\(= 3^{7 – 7} \ \ \ \ (∵a^m ÷ a^n = a{m – n}]\)
\(= 3^0\)
\(= 1\)
(vi) \(2^0 + 3^0 + 4^0\)
Solution:-
\(= 1 + 1 + 1\)
\(= 3\)
(vii) \(2^0 × 3^0 × 4^0\)
Solution:-
\(= 1 × 1 × 1\)
\(= 1\)
(viii) \((3^0 + 2^0) × 5^0\)
Solution:-
= (1 + 1) × 1
= (2) × 1
= 2
(ix) \((2^8 × a^5)\over(4^3 × a^3)\)
Solution:-
\((4)^3 = (2 × 2)^3= (2^2)^3\)
\((2^2)^3 = (2)^{2 × 3}\ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= 2^6\)
\(= {{(2^8 × a^5)}\over {(2^6 × a^3)}}\)
\(= 2^{8 – 6} × a^{5 – 3}\ \ \ \ (∵a^m ÷ a^n = a{m – n}]\)
\(= 2^2 × a^2 \ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= {2a}^2\)
(x) \(({a^5\over a^3}) × a^8\)
Solution:-
\(= (a^{5 -3}) × a^8 \ \ \ \ (∵a^m ÷ a^n = a{m – n}]\)
\(= a^2 × a^8\)
\(= a^{2 + 8} \ \ \ \ (∵a^m × a^n = a^{m + n}]\)
\(= a^{10}\)
(xi) \((4^5 × a^8b^3)\over (4^5 × a^5b^2)\)
Solution:-
\(= 4^{5 – 5} × (a^{8 – 5} × b^{3 – 2}) \ \ \ \ (∵a^m ÷ a^n = a{m – n}]\)
\(= 4^0 × (a^3b)\)
\(= 1 × a^3b\)
\(= a^3b\)
(xii)\((2^{3 × 2})^2\)
Solution:-
\(= (2^{3 + 1})^2 \ \ \ \ (∵a^m × a^n = a^{m + n}]\)
\(= (2^4)^2 = (2)^{4 × 2} \ \ \ \ \ \ \ ( \because (a^m)^n = a^{mn}\)
\(= 2^8\)
3. Say true or false and justify your answer:
(i) 10 × 1011 = 10011
Solution:-
LHS = 10 × 1011
= 101 + 11 … [∵am × an = am + n]
= 1012
RHS = 10011
= (10 × 10)11
= (101 + 1)11
= (102)11
= (10)2 × 11 … [∵(am)n = amn]
= 1022
LHS ≠ RHS
Hence, the given statement is false.
(ii) 23 > 52
Solution:-
LHS = 23= 2 × 2 × 2 = 8
RHS = 52 = 5 × 5 = 25
LHS < RHS
23 < 52
Hence, the given statement is false.
(iii) 23 × 32 = 65
Solution:-
LHS = 23 × 32= 2 × 2 × 2 × 3 × 3 = 72
RHS = 65 = 6 × 6 × 6 × 6 × 6 = 7776
By comparing LHS and RHS,
72 ≠ 7776
LHS ≠ RHS
Hence, the given statement is false.
(iv) 30 = (1000)0
Solution:-
LHS = 30 = 1
RHS = 10000 = 1
LHS = RHS
30 = 10000
Hence, the given statement is true.
4. Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
Solution:-
The factors of 108 = 2 × 2 × 3 × 3 × 3 = 22 × 33
The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 26 × 3
108 × 192= (22 × 33) × (26 × 3)
= 22 + 6 × 33 + 1 … [∵am × an = am + n]
= 28 × 34
(ii) 270
Solution:-
270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64
729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26
729 × 64= 36 × 26
(iv) 768
Solution:-
768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3
5. Simplify:
(i) \(((2^5)^2 × 7^3)\over (8^3 × 7)\)
Solution:-
83 = (2 × 2 × 2)3 = (23)3
\(={ ((2^5)^2 × 7^3)\over ((2^3)^3 × 7)}\)\(= {(2^{5 × 2} × 7^3)\over ((2^{3 × 3} × 7)}\) … [∵(am)n = amn]
\(= (2^{10} × 7^3)\over (2^9 × 7)\)= (210 – 9 × 73 – 1) … [∵am ÷ an = am – n]
= 2 × 72
= 2 × 7 × 7
= 98
(ii) \({(25 × 5^2 \times t^8)\over (10^3 \times t^4)}\)
Solution:-
25 = 5 × 5 = 52
103 = (5 × 2)3 = 53 × 23
\(={ (5^2 × 5^2 × t^8)\over(5^3 × 2^3 × t^4)}\)\(= {(5^{2 + 2} × t^8)\over (5^3 × 2^3 × t^4)} … [∵a^m × a^n = a^{m + n}]\)
\(= {(5^4 × t^8)\over(5^3 × 2^3 × t^4)}\)
\(= {(5^{4 – 3} × t^{8 – 4})\over 2^3} … [∵a^m ÷ a^n = a^{m – n}]\)
\(= {(5 × t^4)\over (2 × 2 × 2)}\)
\(= {(5t^4)\over 8}\)
(iii) \({(3^5 × 10^5 × 25)\over (5^7 × 6^5)}\)
Solution:-
105 = (5 × 2)5 = 55 × 25
25 = 5 × 5 = 52
65 = (2 × 3)5 = 25 × 35
\(= {(3^5 × 5^5 × 2^5 × 5^2)\over (5^7 × 2^5 × 3^5)}\)\(= {(3^5 × 5^{5 + 2} × 2^5)\over (5^7 × 2^5 × 3^5)} … [∵a^m × a^n = a^{m + n}]\)
\(= {(3^5 × 5^7 × 2^5)\over (5^7 × 2^5 × 3^5)}\)
= (35 – 5 × 57 – 7 × 25 – 5)
= (30 × 50 × 20) … [∵am ÷ an = am – n]
= 1 × 1 × 1
= 1
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