DAV Class 7 Maths Chapter 4 Brain Teasers
Exponents and Powers Brain Teasers
1. A. Tick the correct option.
(a) The exponential form of \( \left[ (2^2)^3 \right]^2 \) is- (i) \( 4^5 \) (ii) \( 2^{12} \) (iii) \( (12)^2 \) (iv) \( 2^7 \)
Solution
\[ \begin{align*} &= \left[ (2^2)^3 \right]^2 \\ &= (2^{2 \times 3})^2 \\ &= 2^{6 \times 2} \\ &= 2^{12} \end{align*} \]
Answer \( {\color{orange} (ii)} \ \color{red} 2^{12} \)
(b) \( 2.7 \times 10^{-3} \) is equal to- (i) 0.000027 (ii) 0.00027 (iii) 0.0027 (iv) 2.007
Solution
\[ \begin{align*} &= 2.7 \times 10^{-3} \\ \\ &= 2.7 \times \frac{1}{10^3} \\ \\ &= 2.7 \times \frac{1}{1000} \\ \\ &= 0.0027 \end{align*} \]
Answer \( {\color{orange} (iii)} \ \color{red} 0.0027 \)
(c) \( 9 \times 4^2 \) is same as- (i) \( (12)^2 \) (ii) \( (36)^2 \) (iii) \( (36)^3 \) (iv) \( (18)^4 \)
Solution
\[ \begin{align*} &= 9 \times 4^2 \\ &= 3^2 \times 4^2 \\ &= (3 \times 4)^2 \\ &= (12)^2 \end{align*} \]
Answer \( {\color{orange} (i)} \ \color{red} (12)^2 \)
(d) \( 10 \times 10^{11} \) is equal to- (i) \( (100)^4 \) (ii) \( (10)^{10} \) (iii) \( (100)^{12} \) (iv) \( (10)^{12} \)
Solution
\[ \begin{align*} &= 10 \times 10^{11} \\ &= 10^1 \times 10^{11} \\ &= 10^{1+11} \\ &= 10^{12} \end{align*} \]
Answer \( {\color{orange} (iv)} \ \color{red} (10)^{12} \)
(e) If \( \left( \displaystyle \frac{3}{5} \right)^{2x} = 1 \), then \( x \) is equal to- (i) 2 (ii) 0 (iii) 1 (iv) \( \frac{1}{2} \)
Solution
\[ \begin{align*} \left( \frac{3}{5} \right)^{2x} &= 1 \\ \\ \left( \frac{3}{5} \right)^{2x} &= \left( \frac{3}{5} \right)^0 \\ \\ 2x &= 0 \\ x &= 0 \end{align*} \]
Answer \( {\color{orange} (ii)} \ \color{red} 0 \)
B. Answer the following questions.
(a) Simplify \( \displaystyle \left[\left\{ \left( \frac{4}{9} \right)^2 \right\}^0 \right]^5 \)
\begin{aligned} &= \left[\left\{ \left( \frac{4}{9} \right)^2 \right\}^0 \right]^5 \\ \\ &= \left( \frac{4}{9} \right)^{2 \times 0 \times 5} \\ \\ &= \left( \frac{4}{9} \right)^0 \\ \\ &= 1 \end{aligned}
Answer \( \color{red} 1 \)
(b) Write \( (2^2)^3 \times 3^6 \) in simplified exponential form.
\[ \begin{aligned} &= (2^2)^3 \times 3^6 \\ &= 2^{2 \times 3} \times 3^6 \\ &= 2^6 \times 3^6 \\ &= (2 \times 3)^6 \\ &= 6^6 \end{aligned} \]
Answer \( \color{red} 6^6 \)
(c) Find the value of \( x \) if \( \displaystyle \left[\left( \frac{3}{7} \right)^3 \right]^{-2} = \left( \frac{3}{7} \right)^{2x} \).
\[ \begin{aligned} \left[\left( \frac{3}{7} \right)^3 \right]^{-2} &= \left( \frac{3}{7} \right)^{2x} \\ \\ \left( \frac{3}{7} \right)^{3 \times (-2)} &= \left( \frac{3}{7} \right)^{2x} \\ \\ \left( \frac{3}{7} \right)^{-6} &= \left( \frac{3}{7} \right)^{2x} \\ \\ 2x & = -6 \\ \\ x &= -\frac{\cancel6^3}{\cancel2_1} \\ \\ x &= -3 \end{aligned} \]
Answer \( x =\color{red} -3 \)
(d) Simplify \( \ \displaystyle 6^{-2} + \left( \frac{3}{2} \right)^{-2} \)
\[ \begin{aligned} & = 6^{-2} + \left( \frac{3}{2} \right)^{-2} \\ \\ &= \frac{1}{6^2} + \left( \frac{2}{3} \right)^{2} \\ \\ &= \frac{1}{36} + \frac{4}{9} \\ \\ &= \frac{1}{36} + \frac{4 {\color{green} \times 4}}{9 {\color{green} \times 4}} \\ \\ &= \frac{1+16}{36} \\ \\ &= \frac{17}{36} \end{aligned} \]
Answer \( \displaystyle \color{red} \frac{17}{36} \)
(e) Write in usual form \( 11.2 \times (10)^{-7} \).
\[ \begin{aligned} & = 11.2 \times 10^{-7} \\ \\ &= \frac{11.2}{10^7} \\ \\ &= \frac{11.2}{10000000} \\ \\ &= 0.00000112 \end{aligned} \]
Answer \( \color{red} 0.00000112 \)
2. Write the base and exponent of the following numbers.
(i) \( \displaystyle \left(\frac{1}{9}\right)^{-4}\) \( \implies \displaystyle \text{base} = {\color{red} \frac{1}{9}} , \ \text{exponent} = {\color{red} -4} \)
(ii) \( \displaystyle \frac{7}{11}\) \( \implies \displaystyle \text{base} = {\color{red} \frac{7}{11}} , \ \text{exponent} = {\color{red} 1} \)
(iii) \( \displaystyle \left[\left(\frac{-5}{6}\right)^0\right]^2\) \( \displaystyle \left(\frac{-5}{6}\right)^0\implies \text{base} = {\color{red} \frac{-5}{6}} , \ \text{exponent} = {\color{red} 0} \)
(iv) \( \displaystyle \left(\frac{ac}{b}\right)^3\) \( \implies \displaystyle \text{base} = {\color{red} \frac{ac}{b}} , \ \text{exponent} = {\color{red} 3} \)
(v) \( \displaystyle \left(\frac{-1}{2} \times \frac{1}{3}\right)^0\) \( \displaystyle \left(\frac{-1}{6}\right)^0 \implies \text{base} = {\color{red} \frac{-1}{6}} , \ \text{exponent} = {\color{red} 0} \)
(vi) \( \displaystyle \left(\frac{-11}{12}\right)^4\) \( \implies \displaystyle \text{base} = {\color{red} \frac{-11}{12}} , \ \text{exponent} = {\color{red} 4} \)
3. Express the following as rational number.
(i) \( \displaystyle \left[ \left(-\frac{1}{2}\right)^{-3} \right]^2 \)
\begin{align*} & = \left[ \left(-\frac{1}{2}\right)^{-3} \right]^2 \\ \\ & = \left(-\frac{1}{2}\right)^{-3 \times 2} \\ \\ & = \left(-\frac{1}{2}\right)^{-6} \\ \\ & = \left(\frac{-2}{1}\right)^6 \\ \\ & = \frac{64}{1} \\ \\ & = \color{red} 64 \end{align*}
(ii) \( \displaystyle \left( \frac{1}{3^2} \right)^{-1} \times 9^{-2} \)
\begin{align*} & = \left( \frac{1}{3^2} \right)^{-1} \times 9^{-2} \\ \\ & = 3^{2} \times 9^{-2} \\ & = 9 \times 9^{-2} \\ & = 9^{1-2} \\ & = 9^{-1} \\ & = \color{red} \frac{1}{9} \end{align*}
(iii) \( \displaystyle \left( \frac{16}{25} \right)^{-1} \div \left( \frac{4}{5} \right)^{-3} \)
\begin{align*} & = \left( \frac{16}{25} \right)^{-1} \div \left( \frac{4}{5} \right)^{-3} \\ \\ & = \frac{25}{16} \div \left( \frac{5}{4} \right)^{3} \\ \\ & = \frac{5^2}{4^2} \div \left( \frac{5}{4} \right)^{3} \\ \\ & = \left(\frac{5}{4}\right)^2 \div \left( \frac{5}{4} \right)^{3} \\ \\ & = \left(\frac{5}{4}\right)^{2-3} \\ \\ & = \left(\frac{5}{4}\right)^{-1} \\ \\ & = \color{red} \frac{4}{5} \end{align*}
(iv) \( \displaystyle \left( \frac{2}{3} \right)^{-1} - \left( \frac{3}{2} \right)^{-2} \)
\begin{align*} & = \left( \frac{2}{3} \right)^{-1} - \left( \frac{3}{2} \right)^{-2} \\ \\ & = \frac{3}{2} - \left( \frac{2}{3} \right)^2 \\ \\ & = \frac{3}{2} - \frac{4}{9} \\ \\ & = \frac{3 {\color{green} \times 9}}{2 {\color{green} \times 9}} - \frac{4 {\color{green} \times 2}}{9 {\color{green} \times 2}} \\ \\ & = \frac{27-8}{18} \\ \\ & = \color{red} \frac{19}{18} \end{align*}
(v) \( \displaystyle \left[ \left( \frac{5}{6} \right)^0 + \left( \frac{3}{4} \right)^0 \right] \div \left( \frac{2}{3} \right)^0 \)
\begin{align*} & = \left[ \left( \frac{5}{6} \right)^0 + \left( \frac{3}{4} \right)^0 \right] \div \left( \frac{2}{3} \right)^0 \\ \\ & = \left[ 1 + 1 \right] \div 1 \\ & = 2 \div 1 \\ & = \color{red} 2 \end{align*}
(vi) \( \displaystyle \frac{\left( \frac{-3}{5} \right)^7}{\left( \frac{-3}{5} \right)^7} \)
\begin{align*} & = \frac{\left( \frac{-3}{5} \right)^7}{\left( \frac{-3}{5} \right)^7} \\ \\ & = \left( \frac{-3}{5} \right)^7 \div \left( \frac{-3}{5} \right)^7 \\ \\ & = \left( \frac{-3}{5} \right)^{7-7} \\ \\ & = \left( \frac{-3}{5} \right)^0 \\ \\ & = \color{red} 1 \end{align*}
4. Express the following in exponential form.
(i) \( \displaystyle (2.5)^4 \div \left( \frac{1}{2.5} \right)^2 \)
\begin{align*} & = (2.5)^4 \div \left( \frac{1}{2.5} \right)^2 \\ &= (2.5)^4 \div (2.5)^{-2} \\ &= (2.5)^{4 - (-2)} \\ &= (2.5)^{4 + 2} \\ &= \color{red} (2.5)^6 \end{align*}
(ii) \( \displaystyle (5^2 \times 5^5) \div 5^{10} \)
\begin{align*} & = (5^2 \times 5^5) \div 5^{10} \\ &= 5^{2+5} \div 5^{10} \\ &= 5^{7} \div 5^{10} \\ &= 5^{7 - 10} \\ &= \color{red} 5^{-3} \end{align*}
5. Find the reciprocals.
(i) \( \displaystyle \left( \frac{2}{3} \right)^{-2} \div \left( \frac{2}{3} \right)^3 \)
\begin{align*} & = \left( \frac{2}{3} \right)^{-2} \div \left( \frac{2}{3} \right)^3 \\ \\ & = \left( \frac{2}{3} \right)^{-2-3} \\ \\ & = \left( \frac{2}{3} \right)^{-5} \\ \\ & = \left( \frac{3}{2} \right)^{5} \\ \\ \text{Reciprocal} & = \color{red} \left( \frac{2}{3} \right)^{5} \end{align*}
(ii) \( \displaystyle \left( \frac{3}{4} \right)^{-2} \times \left( \frac{3}{4} \right)^3 \)
\begin{align*} & = \left( \frac{3}{4} \right)^{-2+3} \\ \\ & = \left( \frac{3}{4} \right)^{1} \\ \\ \text{Reciprocal} & = \color{red} \left( \frac{4}{3} \right)^{1} \end{align*}
(iii) \( \displaystyle \left[ \left( \frac{7}{8} \right)^{-3} \right]^2 \)
\begin{align*} & = \left[ \left( \frac{7}{8} \right)^{-3} \right]^2 \\ \\ & = \left( \frac{7}{8} \right)^{-3 \times 2} \\ \\ & = \left( \frac{7}{8} \right)^{-6} \\ \\ & = \left( \frac{8}{7} \right)^{6} \\ \\ \text{Reciprocal} & = \color{red} \left( \frac{7}{8} \right)^{6} \end{align*}
(iv) \( \displaystyle \left( \frac{3}{5} \times \frac{-2}{9} \right)^{-2} \)
\begin{align*} & = \left( \frac{\cancel3^1}{5} \times \frac{-2}{\cancel9_3} \right)^{-2} \\ \\ & = \left( \frac{-2}{5 \times 3} \right)^{-2} \\ \\ & = \left( \frac{-2}{15} \right)^{-2} \\ \\ & = \left( \frac{-15}{2} \right)^{2} \\ \\ \text{Reciprocal} & = \color{red} \left( \frac{-2}{15} \right)^{2} \end{align*}
(v) \( \displaystyle \frac{ \left( \frac{-3}{2} \right)^{-1} \times \left( \frac{2}{3} \right)^2 }{ \left( \frac{2}{3} \right)^{-1} \div \left( \frac{3}{2} \right) } \)
\begin{align*} & = \frac{ \left( \frac{-3}{2} \right)^{-1} \times \left( \frac{2}{3} \right)^2 }{ \left( \frac{2}{3} \right)^{-1} \div \left( \frac{3}{2} \right) } \\ \\ & = \frac{ \left( \frac{-2}{3} \right) \times \left( \frac{4}{9} \right) }{ \left( \frac{3}{2} \right) \div \left( \frac{3}{2} \right) } \\ \\ & = \frac{ \left( \frac{-8}{27} \right) }{ 1 } \\ \\ & = \frac{-8}{27} \\ \\ \text{Reciprocal} & = \color{red} \frac{-27}{8} \end{align*}
6. Express \( \displaystyle \left(15^3 \right)^{-16} \) as a single exponent of 15.
\begin{align*} & = \left(15^3 \right)^{-16} \\ & = 15^{3 \times (-16)} \\ & = 15^{(-48)} \\ &= \color{red} 15^{-48} \end{align*}
7. By what number should \( (7)^{-2} \) be multiplied so that the product may be equal to \( (343)^{-1} \)?
Solution
\begin{align*} \text{Let } x & \text{ be multiplied to } (7)^{-2} \\ x \times 7^{-2} &= 343^{-1} \\ x &= 343^{-1} \div 7^{-2} \\ x &= (7^3)^{-1} \div 7^{-2} \\ x &= 7^{-3} \div 7^{-2} \\ x &= 7^{-3-(-2)} \\ x &= 7^{-3+2} \\ x &= 7^{-1} \\ x &= \frac{1}{7} \end{align*}
Answer The number is \( \color{red} \displaystyle \frac{1}{7} \)
8. By what number should \( \displaystyle \left( \frac{-3}{4} \right)^5 \) be multiplied so that the product may be equal to \( \left( \displaystyle \frac{-64}{27} \right)^{-1} \)?
Solution
\begin{align*} \text{Let } x & \text{ be multiplied to } \left( \frac{-3}{4} \right)^5 \\ \\ x \times \left( \frac{-3}{4} \right)^5 &= \left( \frac{-64}{27} \right)^{-1} \\ \\ x &= \left( \frac{-64}{27} \right)^{-1} \div \left( \frac{-3}{4} \right)^5 \\ \\ x &= \frac{-27}{64} \div \left( \frac{-3}{4} \right)^5 \\ \\ x &= \frac{-3^3}{4^3} \div \left( \frac{-3}{4} \right)^5 \\ \\ x &= \left( \frac{-3}{4} \right)^3 \div \left( \frac{-3}{4} \right)^5 \\ \\ x &= \left( \frac{-3}{4} \right)^{3-5} \\ \\ x &= \left( \frac{-3}{4} \right)^{-2} \\ \\ x &= \left( \frac{-4}{3} \right)^{2} \\ \\ x &= \frac{16}{9} \end{align*}
Answer The number is \( \color{red} \displaystyle \frac{16}{9} \)
9. By what number should \( (-512)^{-1} \) be divided so that the quotient may be equal to \( (8)^{-2} \)?
Solution
\begin{align*} \text{Let } (-512)^{-1} & \text{ be divided by } x \\ \\ (-512)^{-1} \div x &= (8)^{-2} \\ \\ (-512)^{-1} &= (8)^{-2} \times x \\ \\ \frac{-1}{512} &= \frac{1}{8^2} \times x \\ \\ \frac{-1}{8^3} &= \frac{1}{8^2} \times x \\ \\ \frac{-1}{8^3} \div \frac{1}{8^2} &= x \\ \\ \frac{-1}{8^{3-2}} &= x \\ \\ \frac{-1}{8^1} &= x \\ x &= \frac{-1}{8} \end{align*}
Answer The number is \( \displaystyle \color{red} \frac{-1}{8} \)
10. By what number should \( \displaystyle \left( \frac{144}{225} \right)^{-1} \) be divided so that the quotient may be equal to \( \left( \displaystyle \frac{12}{15} \right)^{-4} \)?
Solution
\begin{align*} \text{Let } \left( \frac{144}{225} \right)^{-1} & \text{ be divided by } x \\ \\ \left( \frac{144}{225} \right)^{-1} \div x &= \left( \frac{12}{15} \right)^{-4} \\ \\ \frac{225}{144} \div x &= \left( \frac{15}{12} \right)^{4} \\ \\ \frac{225}{144} \div \left( \frac{15}{12} \right)^{4} &= x \\ \\ \frac{15^2}{12^2} \div \left( \frac{15}{12} \right)^{4} &= x \\ \\ \left( \frac{15}{12} \right)^{2} \div \left( \frac{15}{12} \right)^{4} &= x \\ \\ \left( \frac{15}{12} \right)^{2-4} &= x \\ \\ \left( \frac{15}{12} \right)^{-2} &= x \\ \\ \left( \frac{12}{15} \right)^{2} &= x \\ \\ \frac{144}{225} &= x \\ \\ x & = \frac{144}{225} \end{align*}
Answer The number is \( \displaystyle \color{red} \frac{144}{225} \)
11. Write the following numbers in the usual form.
(i) \( 5.3 \times 10^5 \)
Solution
\begin{align*} &= 5.3 \times 10^5 \\ &= 5.3 \times 100000 \\ &= \color{red} 530000 \end{align*}
(ii) \( 2.9 \times 10^{-10} \)
Solution
\begin{align*} &= 2.9 \times 10^{-10} \\ &= 2.9 \times \frac{1}{10000000000} \\ \\ &= \color{red} 0.00000000029 \end{align*}
(iii) \( 4.6 \times 10^{-12} \)
Solution
\begin{align*} &= 4.6 \times 10^{-12} \\ &= 4.6 \times \frac{1}{1000000000000} \\ &= \color{red} 0.0000000000046 \end{align*}
(iv) \( 1.08 \times 10^7 \)
Solution
\begin{align*} &= 1.08 \times 10^7 \\ &= 1.08 \times 10000000 \\ &= \color{red} 10800000 \end{align*}
(v) \( 3.09 \times 10^{11} \)
Solution
\begin{align*} &= 3.09 \times 10^{11} \\ &= 3.09 \times 100000000000 \\ &= \color{red} 309000000000 \end{align*}
(vi) \( 6.00005 \times 10^9 \)
Solution
\begin{align*} &= 6.00005 \times 10^9 \\ &= 6.00005 \times 1000000000 \\ &= \color{red} 6000050000 \end{align*}
12. Write the following numbers in the form \( k \times 10^n \) where \( 1 \leq k < 10 \) and \( n \) is an integer.
(i) \( 762850 \)
Solution
\begin{align*} & = 762850 \\ &= \color{red} 7.6285 \times 10^5 \end{align*}
(ii) \( 2500000 \)
Solution
\begin{align*} & = 2500000 \\ &= \color{red} 2.5 \times 10^6 \end{align*}
(iii) \( 0.09 \)
Solution
\begin{align*} & = 0.09 \\ &= \color{red} 9.0 \times 10^{-2} \end{align*}
(iv) \( 0.0000076 \)
Solution
\begin{align*} & = 0.0000076 \\ &= \color{red} 7.6 \times 10^{-6} \end{align*}
(v) \( 0.00000008 \)
Solution
\begin{align*} & = 0.00000008 \\ &= \color{red} 8.0 \times 10^{-8} \end{align*}
(vi) \( 4592000000000 \)
Solution
\begin{align*} & = 4592000000000 \\ &= \color{red} 4.592 \times 10^{12} \end{align*}
(vii) \( \displaystyle \frac{315 \times 10^5}{0.7 \times 10^3} \)
Solution
\begin{align*} & = \frac{315 \times 10^5}{0.7 \times 10^3} \\ \\ & = \frac{\cancel{315}^{45} \times 10^5}{\cancel 7_1 \times 10^2} \\ \\ & = 45 \times 10^{5-2} \\ & = 45 \times 10^{3} \\ &= \color{red} 4.5 \times 10^4 \end{align*}
(viii) \( \displaystyle \frac{4.4 \times 10^{-7}}{44 \times 10^{-5}} \)
Solution
\begin{align*} & = \frac{4.4 \times 10^{-7}}{44 \times 10^{-5}} \\ \\ & = \frac{\cancel{44}^1 \times 10^{-8}}{\cancel{44}_1 \times 10^{-5}} \\ \\ &= 1 \times 10^{-8-(-5)} \\ &= 1 \times 10^{-8+5} \\ &= \color{red} 1.0 \times 10^{-3} \end{align*}
(ix) \( ( 7 \times 10^{2}) \times (8 \times 10^3) \)
Solution
\begin{align*} & = ( 7 \times 10^{2}) \times (8 \times 10^3) \\ & = 7 \times 8 \times 10^{2} \times 10^3 \\ & = 56 \times 10^{2+3} \\ & = 56 \times 10^{5} \\ & = \color{red} 5.6 \times 10^{6} \end{align*}
13. Simplify the following:
(i) \( \displaystyle \left( \frac{7}{8} \right)^{-3} \times \left( \frac{9}{5} \right)^0 \times 8^{-2} \times \left( \frac{1}{7} \right)^{-1} \)
Solution
\begin{align*} & = \left( \frac{7}{8} \right)^{-3} \times \left( \frac{9}{5} \right)^0 \times 8^{-2} \times \left( \frac{1}{7} \right)^{-1} \\ \\ &= \left( \frac{8}{7} \right)^3 \times 1 \times \frac{1}{8^2} \times 7^1 \\ \\ &= \frac{8^3}{7^3} \times \frac{1}{8^2} \times 7^1 \\ \\ &= \frac{8^{3-2}}{7^{3-1}} \\ \\ &= \frac{8^1}{7^2} \\ \\ &= \color{red} \frac{8}{49} \end{align*}
(ii) \( \displaystyle \left\{ \left[ \left( \frac{4}{5} \right)^3 \right]^2 \div \left( \frac{4}{5} \right)^2 \right\} \times \left( \frac{1}{4} \right)^{-2} \times 4^{-1} \)
Solution
\begin{align*} & = \left\{ \left[ \left( \frac{4}{5} \right)^3 \right]^2 \div \left( \frac{4}{5} \right)^2 \right\} \times \left( \frac{1}{4} \right)^{-2} \times 4^{-1} \\ \\ &= \left\{ \left( \frac{4}{5} \right)^{3 \times 2} \div \left( \frac{4}{5} \right)^2 \right\} \times 4^{2} \times 4^{-1} \\ \\ &= \left\{ \left( \frac{4}{5} \right)^{6} \div \left( \frac{4}{5} \right)^2 \right\} \times 4^{2+(-1)} \\ \\ &= \left( \frac{4}{5} \right)^{6-2} \times 4^{2-1} \\ \\ &= \left( \frac{4}{5} \right)^4 \times 4^1 \\ \\ &= \frac{4^4}{5^4} \times 4 \\ \\ &= \frac{4^{4+1}}{5^4} \\ \\ &= \frac{4^5}{5^4} \\ \\ &= \color{red} \frac{1024}{625} \end{align*}
(iii) \( \displaystyle \left( \frac{-4}{5} \right)^3 \times 5^2 \times \left( \frac{-1}{2} \right)^5 \times \left( \frac{1}{2} \right)^{-3} \)
Solution
\begin{align*} & = \left( \frac{-4}{5} \right)^3 \times 5^2 \times \left( \frac{-1}{2} \right)^5 \times \left( \frac{1}{2} \right)^{-3} \\ \\ &= \frac{(-4)^3}{5^3} \times 5^2 \times \frac{(-1)^5}{2^5} \times 2^3 \\ \\ &= \frac{(-4)^3}{5^{3-2}} \times \frac{-1}{2^{5-3}} \\ \\ &= \frac{-64}{5^{1}} \times \frac{-1}{2^{2}} \\ \\ &= \frac{\cancel{64}^{16}}{5} \times \frac{1}{\cancel4_1} \\ \\ &= \color{red} \frac{16}{5} \end{align*}
(iv) \( \displaystyle \left[ \left( \frac{2}{3} \right)^2 \right]^{-6} \times \left( \frac{2}{3} \right)^2 \times \left( \frac{3}{2} \right)^{-10} \)
Solution
\begin{align*} & = \left[ \left( \frac{2}{3} \right)^2 \right]^{-6} \times \left( \frac{2}{3} \right)^2 \times \left( \frac{3}{2} \right)^{-10} \\ \\ &= \left( \frac{2}{3} \right)^{-12} \times \left( \frac{2}{3} \right)^2 \times \left( \frac{2}{3} \right)^{10} \\ \\ &= \left( \frac{2}{3} \right)^{-12+2+10} \\ \\ &= \left( \frac{2}{3} \right)^{-12+12} \\ \\ &= \left( \frac{2}{3} \right)^0 \\ \\ &= \color{red} 1 \end{align*}
14. Express the numbers appearing in the following statements in the form \( k \times 10^n \) where \( 1 \leq k < 10 \) and \( n \) is an integer.
(i) The mean distance of the moon from the earth is 384,400,000 metres.
Solution
\begin{align*} &= 384,400,000 \ metres \\ &= 3.844 \times 10^8 \ metres \\ \end{align*}
Answer \( \color{red} 3.844 \times 10^8 \ metres\)
(ii) The distance travelled by a ray of light in one year is 9,460,500,000,000,000 m.
Solution
\begin{align*} &= 9,460,500,000,000,000 \ metres \\ &= 9.4605 \times 10^{15} \ metres \\ \end{align*}
Answer \( \color{red} 9.4605 \times 10^{15} \ metres \)
(iii) The number of red blood cells per cubic mm of human blood is approximately \( 5.5 \) million.
Solution
\begin{align*} &= 5.5 \times 1,000,000 \\ &= 5.5 \times 10^6 \\ \end{align*}
Answer \( \color{red} 5.5 \times 10^6 \) cells per cubic mm
15. Find the value of \( x \) so that:
(i) \( \displaystyle \left( \frac{-7}{11} \right)^{-3} \times \left( \frac{-7}{11} \right)^{5x} = \left[ \left( \frac{-7}{11} \right)^{-2} \right]^{-1} \)
Solution
\begin{align*} \left( \frac{-7}{11} \right)^{-3} \times \left( \frac{-7}{11} \right)^{5x} &= \left[ \left( \frac{-7}{11} \right)^{-2} \right]^{-1} \\ \\ \left( \frac{-7}{11} \right)^{-3 + 5x} &= \left( \frac{-7}{11} \right)^{2} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ -3 + 5x &= 2 \\ 5x &= 2 + 3 \\ 5x &= 5 \\ x &= \frac{\cancel5^1}{\cancel 5_1} \\ \color{red} x &= \color{red}1 \end{align*}
(ii) \( \displaystyle \left( \frac{3}{7} \right)^{-2x+1} \div \left( \frac{3}{7} \right)^{-1} = \left[ \left( \frac{3}{7} \right)^{-1} \right]^{-7} \)
Solution
\begin{align*} \left( \frac{3}{7} \right)^{-2x+1} \div \left( \frac{3}{7} \right)^{-1} &= \left( \frac{3}{7} \right)^{7} \\ \\ \left( \frac{3}{7} \right)^{-2x+1 - (-1)} &= \left( \frac{3}{7} \right)^{7} \\ \\ \left( \frac{3}{7} \right)^{-2x+1+1} &= \left( \frac{3}{7} \right)^{7} \\ \\ \left( \frac{3}{7} \right)^{-2x+2} &= \left( \frac{3}{7} \right)^{7} \\ \\ \text{Since bases are same, } & \text{their exponents must be equal.} \\ -2x+2 &= 7 \\ -2x &= 7 - 2 \\ -2x &= 5 \\ \\ -x &= \frac{5}{2} \\ \\ \color{red} x &= \color{red} \frac{-5}{2} \end{align*}
16. If \( \displaystyle \frac{p}{q} = \left( \frac{5}{6} \right)^{-2} \times \left( \frac{4}{3} \right)^{0} \), find the value of \( \displaystyle \left( \frac{p}{q} \right)^{-2} \)
Solution
\begin{align*} \frac{p}{q} &= \left( \frac{5}{6} \right)^{-2} \times \left( \frac{4}{3} \right)^0 \\ \\ &= \left( \frac{6}{5} \right)^2 \times 1 \\ \\ &= \frac{36}{25} \\ \\ \color{green} \frac{p}{q} &= \color{green} \frac{36}{25} \\ \\ \left( \frac{p}{q} \right)^{-2} &= \left( \frac{36}{25} \right)^{-2} \\ \\ &= \left( \frac{25}{36} \right)^2 \\ \\ &= \frac{625}{1296} \\ \\ \color{red} \left( \frac{p}{q} \right)^{-2} &= \color{red} \frac{625}{1296} \end{align*}