Q.2 Exress the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c× c × d
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \( 6 × 6 × 6 × 6 = 6^3 \)
(ii)\( t × t = t^2 \)
(iii) \( b × b × b × b = b^4 \)
(iv) \(5 × 5× 7 × 7 × 7 = 5^2 × 7^3 = 5^2 · 7^3\)
(v) \(2 × 2 × a × a = 2^2 × a^2 = 2^2 · a^2 \)
(vi)\( a × a ×a × c × c × c × c × d = a^3 × c^4 × d = a^3 · c^4 · d \)
Q.3 Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) $$
\begin{array}{|l}
\llap{2~~~~} 512 \\ \hline
\llap{2~~~~} 216 \\ \hline
\llap{2~~~~} 128 \\ \hline
\llap{2~~~~} 64 \\ \hline
\llap{2~~~~} 32 \\ \hline
\llap{2~~~~} 16 \\ \hline
\llap{2~~~~} 8 \\ \hline
\llap{2~~~~} 4 \\ \hline
\llap{2~~~~} 2 \\ \hline
\llap{~~~~} 1 \\ \hline
\end{array}
$$
512= 2×2×2×2×2×2×2×2×2 = \(2^9\)
(ii) $$
\begin{array}{|l}
\llap{7~~~~} 343 \\ \hline
\llap{7~~~~} 49 \\ \hline
\llap{7~~~~} 7 \\ \hline
\llap{~~~~} 1 \\ \hline
\end{array}
$$
343 = 7×7×7 = \(7^3\)
(iii) $$
\begin{array}{|l}
\llap{3~~~~} 729 \\ \hline
\llap{3~~~~} 243 \\ \hline
\llap{3~~~~} 81 \\ \hline
\llap{3~~~~} 27 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{~~~~} 3 \\ \hline
\end{array}
$$
729 = 3×3×3×3×3×3 = \(3^6\)
(iv) $$
\begin{array}{|l}
\llap{5~~~~} 3125 \\ \hline
\llap{5~~~~} 625 \\ \hline
\llap{5~~~~} 125 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
\llap{~~~~} 1 \\ \hline
\end{array}
$$
3125 = 5×5×5×5×5= \(5^5 \)
Q.4 Identify the greater number, wherever possible, in each of the following?
(i) \( 4^3 or 3^4 \)
(ii)\( 5^3 or 3^5 \)
(iii)\( 2^8 or 8^2 \)
(iv)\( 100^2 or 2^{100} \)
(v) \(2^{10} or 10^2 \)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \( 4^3 or 3^4 \)
\(4^3 = 4 × 4 × 4 = 64, \)
\(3^4 = 3 × 3 × 3 × 3 = 81\)
Since 81 > 64
\(\therefore 3^4 > 4^3. \)
(ii)\( 5^3 or 3^5 \)
\(5^3 = 5 × 5 × 5 = 125 \)
\(3^5 = 3 × 3 × 3 × 3 × 3 = 243 \)
Since 243 > 125
\( 3^5 > 5^3. \)
(iii)\( 2^8 or 8^2 \)
\(2^8 =2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256\)
\(8^2 = 8 × 8 = 64 \)
Since 256 > 64
\( 2^8 > 8^2.\)
(iv) \(100^2 or 2^{100}\)
\(100^2 = 100 × 100 = 10000 \)
\(2^{100} = 2 × 2 × 2 × … 100 times = 214 = 16384 \)
Since 16384 > 10,000
\( 2^{100} > 100^2. \)
(v)\( 2^{10} or 10^2 \)
\(2^{10} = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 \)
\(10^2 = 10 × 10 = 100 \)
\(\therefore 2^{10} >10^2. \)
Q.5 Express each of the following as the product of powers of their prime
(i) 648
(ii) 405
(iii) 540
(iv) 3600
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) $$
\begin{array}{|l}
\llap{2~~~~} 648 \\ \hline
\llap{2~~~~} 324 \\ \hline
\llap{2~~~~} 162 \\ \hline
\llap{3~~~~} 81 \\ \hline
\llap{3~~~~} 27 \\ \hline
\llap{3~~~~} 9 \\ \hline
\llap{3~~~~} 3 \\ \hline
1
\end{array}
$$
648= \(2^3× 3^4 \)
(ii) $$
\begin{array}{|l}
\llap{3~~~~} 405 \\ \hline
\llap{3~~~~} 135 \\ \hline
\llap{3~~~~} 45 \\ \hline
\llap{3~~~~} 15 \\ \hline
\llap{5~~~~} 5 \\ \hline
\llap{~~~~} 1 \\ \hline
\end{array}
$$
405= \(3^4×5^1 \)
(iii) $$
\begin{array}{|l}
\llap{2~~~~} 540 \\ \hline
\llap{2~~~~} 270 \\ \hline
\llap{3~~~~} 135 \\ \hline
\llap{3~~~~} 45 \\ \hline
\llap{3~~~~} 15 \\ \hline
\llap{5~~~~} 5 \\ \hline
\llap{~~~~} 1 \\ \hline
\end{array}
$$
\( 540 = 2^2×3^3×5^1 \)
(iv) $$
\begin{array}{|l}
\llap{2~~~~} 3600 \\ \hline
\llap{2~~~~} 1800 \\ \hline
\llap{2~~~~} 900 \\ \hline
\llap{2~~~~} 450 \\ \hline
\llap{3~~~~} 225 \\ \hline
\llap{3~~~~} 75 \\ \hline
\llap{5~~~~} 25 \\ \hline
\llap{5~~~~} 5 \\ \hline
1
\end{array}
$$
\(3600 = 2^4×3^2×5^2 \)
Q.6 Simplify:
(i) \(2 × 10^3\)
(ii) \(7^2 × 2^2\)
(iii) \(2^3 × 5\)
(iv) \(3 × 4^4\)
(v) \(0 × 10^2\)
(vi)\( 5^2 × 3^3\)
(vii) \(2^4 × 3^2\)
(viii)\( 3^2 × 10^4\)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \(2 × 10^3 = 2 × 10 × 10 × 10 = = 2000\)
(ii)\( 7^2 × 2^2 = = 7 × 7 × 2 × 2 = 196\)
(iii)\( 2^3 × 5 = 2 × 2 × 2 × 5 = 40\)
(iv) \(3 × 4^4 = 3 × 4 × 4 × 4 × 4 = 768\)
(v)\( 0 × 10^2 = 0 × 10 × 10 = = 0\)
(vi) \(5^2 × 3^3 = 5 × 5 × 3 × 3 × 3 = 675\)
(vii) \(2^4 × 3^2 = 2 × 2 × 2 × 2 × 3 × 3 = 144\)
(viii) \(3^2 × 10^4 = 3 × 3 × 10 × 10 × 10 × 10 = 90000\)
Q.7 Simplify:
(i)\( (-4)^3\)
(ii)\( (-3) × (-2)^3\)
(iii)\( (-3)^2 × (-5)^2\)
(iv)\( (-2)^3 × (-10)^3\)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i)\( (-4)^3 = (-4) × (-4) × (-4) = -64\) [\(\therefore \)(-a)odd number = (-a)odd number]
(ii) \( (-3) × (-2)^3 = (-3) × (-2) × (-2) × (-2)\)
\(= (-3) × (-8) \)= 24 [\(\therefore \)(-a)odd number = (-a)odd number]
(iii)\( (-3)^2 × (-5)^2 = [(-3) × (-5)]^2\)
\(= 15^2 = 225 [\therefore a^m × b^m = (ab)^m] \)
(iv)\( (-2)^3 × (-10)^3 = [(-2) × (-10)]^3\)
\(= 20^2 = 8000 [\therefore a^m × b^m = (ab)^m]\)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \( 2.7 × 10^{12} = 2.7 × 10^4×10^8 = 27000×10^8\)
\(27000×10^8 > 1.5 × 10^8\)
\( 2.7 × 10^{12} > 1.5 × 10^8 \)
(ii) \( 3× 10^{17} = 3× 10^3 ×10^{14} = 3000× 10^{14} \)
\(3000× 10^{14} > 4× 10^{14}\)
\( 4 × 10^{14} < 3 × 10^{17}\)
Q.1 Using laws of e×ponents, simplify and write the answer in e×ponential form:
(i) \(3^2 × 3^4 × 3^8\)
(ii)\( 6^{15} ÷ 6^{10}\)
(iii)\( a^3 × a^2\)
(iv)\( 7^x × 7^2\)
(v)\( (5^2)^3 ÷ 5^3\)
(vi) \(2^5 × 5^5\)
(vii) \(a^4 × b^4\)
(viii)\( (3^4)^3\)
(ix)\( (2^{20 }÷ 2^{15}) × 2^3\)
(x)\( 8^t ÷ 8^2\)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \(3^2 × 3^4 × 3^8 = 3^{2+4+8} = 3^{14} [a^m ÷ a^n = a^{m+n}]\)
(ii) \(6^{15} ÷ 6^{10} = 6^{15-10} = 6^5 [a^m ÷ a^n = a^{m-n}]\)
(iii) \(a^3 × a^2 = a^{3+2} = a^5 [a^m × a^n = a^{m+n}]\)
(iv)\( 7^x × 7^2 = 7^{x+2} [a^m × a^n = a^{m+n}]\)
(v)\( (5^2)^3 ÷ 5^3 = 5^{2×3} ÷ 5^3 = 5^6 ÷ 5^3 = 5^{6-3}\)
\(= 5^3 [(a^m)^n = a^{mn}, a^m ÷ a^n = a^{m-n}]\)
(vi) \( 2^5 × 5^5 = (2 × 5)^5 = 10^5 [a^m × b^m = (ab)^m] \)
(vii) \( a^4 × b^4 = (ab)^4 [a^m × b^m = (ab)^4] \)
(ix)\( (2^{20} ÷ 2^{15}) × 2^3 = 2^{20-15} × 2^3
=2^5 × 2^3 = 2^{5+3} = 2^8 \)
(x) \( 8^t ÷ 8^2 = 8^{t-2} \)
Q.2 Simplify and express each of the following in exponential form:
(i) \(\frac{(2^3 × 3^4 × 4)}{(3 × 32)} \)
(ii) \([(5^2)^3 × 5^4]÷ 5^7 \)
(iii) \(25^4÷5^3 \)
(iv) \( \frac{3×7^2×11^8}{21×11^3} \)
(v) \( \frac{3^7}{3^4×3^3} \)
(vi) \( 2^0+3^0+4^0 \)
(vii) \( 2^0×3^0×4^0 \)
(viii) \( (3^0+2^0)×5^0 \)
(ix)\( \frac{2^8×a^5}{4^3×a^3} \)
(x) \( (\frac{a^5}{a^3} ) × a^8 \)
(xi) \( \frac{4^5×a^8b^3}{4^5×a^5b^2} \)
(xii) \((2^3× 2)^2 \)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \(\frac{2^3 × 3^4 × 4}{3 × 32} \) = \(\frac{2^3 × 3^4 × 2^2}{3 × 2^5} \)
= \(\frac{2^{3+2} × 3^4}{3 × 2^5} \)=\(\frac{2^{5} × 3^4}{3 × 2^5} \)
= \( 2^{5-5}×3^{4-1} \) = \( 2^{0}×3^{3} \)= \(3^3\)
(ii)\([(5^2)^3 × 5^4]÷ 5^7 \) = \((5^6 × 5^4)÷ 5^7 \)
= \((5^{6+4} ÷ 5^7 \) = \(5^{10-7} \) = \(5^3\)
(iii) \(25^4÷5^3 \) = \((5^2)^4÷5^3 \)= \(5^8÷5^3 \) = \(5^{8-3} \)= \(5^5 \)
(iv) \( \frac{3×7^2×11^8}{21×11^3} \) = \( \frac{3×7^2×11^8}{7×3×11^3} \)
= \( 7^{2-1}×11^{8-3} \) = \( 7×11^{5} \) \)
(v) \( \frac{3^7}{3^4×3^3} \) = \( \frac{3^7}{3^{4+3}} \) = \( \frac{3^7}{3^7} \) = 1
(vi) \( 2^0+3^0+4^0 \) = 1+1+1 = 0\( [ \therefore a^0= 1 ] \)
(vii) \( 2^0×3^0×4^0 \) = 1×1×1 = 1
(viii) \( (3^0+2^0)×5^0 \) = (1+1)×1=2×1=2
(ix) \( \frac{2^8×a^5}{4^3×a^3} \) = \( \frac{2^8×a^5}{(2^2)^3×a^3} \)
= \( \frac{2^8×a^5}{2^6×a^3} \) = \( 2^{8-6}{×a^{5-3} \) = \( 2^2×a^2 \) = \((2a)^2\)
(x) \( (\frac{a^5}{a^3} ) × a^8 \) = \( \frac{a^{5+8}}{a^3} \)
= \( \frac{a^{13}}{a^3} \)
= \(a^{13-3} \) = \(a^{10} \)
(xi) \( \frac{4^5×a^8b^3}{4^5×a^5b^2} \) = \( \frac{a^8b^3}{a^5b^2} \)
= \(a^{8-5}b^{3-2} \) = \(a^3b\)
(xii) \((2^3× 2)^2 \) = \(2^6 ×2^2 \) = \( 2^{6+2} \) = \(2^{8} \)
Q.3 Say true or false and justify your answer:
(i)\( 10 × 10^11 = 100^11\)
(ii) \( 2^3 > 5^2 \)
(iii)\( 2^3 × 3^2 = 6^5 \)
(iv) \( 3^20 = (1000)^0 \)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i)LHS = \( 10 × 10^11 = 10^{1+11} = 10^{12} \)
RHS\( = 100^11 = (10^2)^11 = 10^22 \)
\( 10^12 \neq 10^22 \)
\(\therefore \) Statement is false.
(ii)\( 2^3 > 5^2 \)
LHS = \(2^3 = 8 \)
RHS =\( 5^2 = 25 \)
8 < 25
\(\therefore 2^3 < 5^2 \)
Thus, the statement is false.
(iii) \(2^3 × 3^2 = 6^5 \)
LHS \(= 2^3 × 3^2 = 8 × 9 = 72 \)
RHS \(= 6^5 = 6 × 6 × 6 × 6 × 6 = 7776 \)
\( 72 \neq 7776 \)
\(\therefore \)The statement is false.
(iv)\( 3^0 = (1000)^0 \)
LHS = \( 3^0 = 1 \)
RHS = \( (1000)^0 = 1 \)
LHS = RHS
Q.4 Express each of the following as a product of prime factors only in exponential form:
(i) 108 × 192
(ii) 270
(iii) 729 × 64
(iv) 768
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) 108 × 192 = 2 × 2 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3
=\(2^8 × 3^4\)
(ii) \( 270 = 2×3×3×3×5 = 2×3^3×5 \)
(iii) 729 × 64 = 3 × 3 × 3 × 3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 2
\(=3^6 × 2^6 \)
(iv) 768 = 2×2×2×2×2×2×2×2×3 = \( 2^8 ×3 \)
Q.5 Simplify:
(i) \(\frac{(2^5)^2× 7^3}{8^3 × 7} \)
(ii) \(\frac{25×5^2× t^8}{10^3 ×t^4 } \)
(iii) \(\frac{3^5×10^5× 25}{5^7 ×6^5 } \)
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) \(\frac{(2^5)^2× 7^3}{8^3 × 7} \) = \(\frac{(2^{10}× 7^3}{(2^3)^3 × 7} \)
= \(\frac{(2^{10}× 7^3}{{2^9 × 7} \) = \(2^{10-9}× 7^{3-1}\) = \(2^1× 7^2\) =2× 49 = 98
(ii) \(\frac{25×5^2× t^8}{10^3 ×t^4 } \) = \(\frac{5^2×5^2× t^8}{2^3×5^3×t^4 } \)
= \(\frac{5^4× t^8}{2^3×5^3 ×t^4 } \) = \(\frac{5^{4-3}× t^{8-4}}{2^3} \) = \(\frac{5t^4}{2^3} \)
(iii) \(\frac{3^5×10^5× 25}{5^7 ×6^5 } \)= \(\frac{3^5×2^5×5^5× 5^2}{5^7 ×2^5×3^5 } \)
= \(\frac{3^5×5^7× 2^5}{5^7 ×2^5×3^5} \) = 1
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i) 279404 = 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 \( = 2 × 10^5 + 7 × 10^4 + 9 × 10^3 + 4 × 10^2 + 0 × 10^1 + 4 × 10^0 \) (ii) 3006194 = 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 \(= 3 × 10^6 + 0 × 10^5 + 0 × 10^4 + 6 × 10^3 + 1 × 10^2 + 9 × 10^1 + 4 × 10^0 \) (iii) 2806196 = 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 6 \( = 2 × 10^6 + 8 × 10^5 + 0 × 10^4 + 6 × 10^3 + 1 × 10^2 + 9 × 10^1 + 6 × 10^0 \) (iv) 120719 = 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 \(= 1 × 10^5 + 2 × 10^4 + 0 × 10^3 + 7 × 10^2 + 1 × 10^1 + 9 × 10^0 \) (v) 20068 = 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 \( = 2 × 10^4 + 0 × 10^3 + 0 × 10^2 + 6 × 10^1 + 8 × 10^0\)
Q.2 Find the number from each of the following expanded forms:
(a) 8 × 10^4 + 6 × 10^3 + 0 × 10^2 + 4 × 10^1 + 5 × 10^0
(b) 4 × 10^5 + 5 × 10^3 + 3 × 10^2 + 2 × 10^0
(c) 3 × 10^4 + 7 × 10^2 + 5 × 10^0
(d) 9 × 10^5 + 2 × 10^2 + 3 × 10^1
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(a)\( 8 × 10^4 + 6 × 10^3 + 0 × 10^2 + 4 × 10^1 + 5 × 10^0\)
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80000 + 6000 + 0 + 40 + 5 = 86045
(b)\( 4 × 10^5 + 5 × 10^3 + 3 × 10^2 + 2 × 10^0\)
= 4 × 100000 + 5 × 1000 + 3 × 100 + 2 × 1
= 400000 + 5000 + 300 + 2 = 405302
(c) \( 3 × 10^4 + 7 × 10^2 + 5 × 10^0 \)
= 3 × 10000 + 7 × 100 + 5 × 1
= 30000 + 700 + 5 = 30705
(d)\( 9 × 10^5 + 2 × 10^2 + 3 × 10^1 \)
= 9 × 100000 + 2 × 100 + 3 × 10
= 900000 + 200 + 30 = 900230
Q. 3 Express the following numbers in standard form:
(i) 5,00,00,000
(ii) 70,00,000
(iii) 3,18,65,00,000
(iv) 3,90,878
(v) 39087.8
(vi) 3908.78
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(i)\( 5,00,00,000 = 5 × 10^7 \)
(ii)\( 70,00,000 = 7 × 10^6 \)
(iii) \( 3,18,65,00,000 = 3.1865 × 10^9 \)
(iv) \(3,90,878 = 3.90878 × 10^5 \)
(v)\( 39087.8 = 3.90878 × 10^4 \)
(vi)\( 3908.7 8 = 3.90878 × 10^3 \)
Q.4 Express the number appearing in the following statements in standard form:
(a) The distance between Earth and Moon is 384.0. 000 m.
(b) Speed of light in vacuum is 300,000,000 m/s.
(c) Diameter of the Earth is 1,27,56,000 m.
(d) Diameter of the Sun is 1,400,000,000 m.
(e) In a galaxy there are an average 100,000,000,000 stars.
(f) The universe is estimated to be about 12,000,000,000 years old.
(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be 300,000,000,000,000,000,000 m.
(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.
(i) The Earth has 1,353,000,000 cubic km of sea water.
(j) The population of India was about 1,027,000,000 in March 2001.
NCERT Solutions for Class 7 Maths Chapter 13 Exponents And Powers
Answer :
(a)\( 384,000,000 m = 3.84 × 10^8 m \)
(b) \( 300,000,000 m/s = 3 × 10^8 m/s \)
(c)\( 1,27,56,000 m = 1.2756 × 10^7 m \)
(d) \( 1,400,000,000 m = 1.4 × 10^9 m \)
(e) \( 100,000,000,000 stars = 1 × 10^{11} stars \)
(f) \( 12,000,000,000 years old = 1.2 × 10^{10} years old \)
(g) \( 300,000,000,000,000,000,000 m = 3 × 10^{20} m \)
(h)\( 60, 230, 000, 000, 000, 000, 000, 000 molecules = 6.023 × 10^{22}molecules\)
(i) \( 1,353,000,000 cubic km = 1.353 × 10^9 cubic km \)
(j)\( 1,0,27,000,000 = 1.027 × 10^9 \)