EXERCISE 4.2
Question 1
As we know that $a^{m} \times a^{n}=a^{m+n}$
i) $2^{7} \times 2^{4}$
$=2^{7+4}=2^{11}$
ii) $p^{5} \times p^{3}=p^{5+3}=p^{8}$
iii) iii. $(-7)^{5} \times(-7)^{n}=(-7)^{5+11}=(-7)^{16}$
iv) $\left(\frac{3}{5}\right)^{6} \div\left(\frac{3}{5}\right)^{2}$
As we know $a^{m} \div a^{n}=a^{m-n}$
i.e $=\left(\frac{3}{5}\right)^{6-2}=\left(\frac{3}{5}\right)^{4}$
v)
$\begin{aligned}(-6)^{7} \div(-6)^{3} \\=&(-6)^{7-3} \\=(-6)^{4} \end{aligned}$
Question 2
i) $5^{3} \times 5^{7} \times 5^{12}$
$=5^{3+7+12}$
$=5^{22}$
ii) $a^{5} \times a^{3} \times a^{7}$
$=a^{5+3+7}$
$=a^{15}$
iii)$\left(7^{12} \times 7^{3}\right) \div 7^{4}$
$=7^{12+3} \div 7^{4}$
$=7^{15} \div 7^{4}$
$=7^{15-4}$
$=7^{11}$
Question 3
i) $\left(2^{2}\right)^{100}$
As we $\left(a^{m}\right)^{n}=a^{m n}$
$=2^{2 \times 100}=2^{200}$
ii) $\left((-7)^{6}\right)^{5}$
$=(-7)^{6 \times 5}=(-7)^{30}$
iii) $\quad\left(3^{2}\right)^{5} \times\left(3^{4}\right)^{7}$
$=3^{2 \times 5} \times 3^{4 \times 7}$
$=3^{10} \times 3^{28}$
$=3^{10+28}$
$=3^{38}$
Question 4
i) $\frac{a^{3} \times a^{5}}{\left(a^{3}\right)^{2}}$
$=\frac{a^{3+5}}{a^{3 x^{2}}}$
$=\frac{a^{8}}{a^{6}}$
$=a^{8-6}=a^{2}$
ii) $\left(2^{3}\right)^{4} \div 2^{5}$
$=2^{3 \times 4} \div 2^{5}$
$=2^{12} \div 2^{5}$
$=2^{12-5}$
$=2^{7}$
iii. $\left[\left(6^{2}\right)^{3} \div 6^{3}\right] \times 6^{5}$
$=\left[6^{2 \times 3}+6^{3}\right] \times 6^{5}$
$=\left[6^{6} \div 6^{3}\right] \times 6^{5}$
$=6^{6-3} \times 6^{5}$
$=6^{3} \times 6^{5}$
$=6^{3+5}$
$=6^{8}$
Question 5
i) $5^{4} \times 8^{4}$
As we know that $a^{m} \times b^{m}=(a \times b)^{m}=(a b)^{m}$
$=(5 \times 8)^{4}$
$=40^{4}$
ii)
$\begin{aligned} &(-3)^{6} \times(-5)^{6} \\ &=(-3 x-5)^{6} \\ &=15^{6} \end{aligned}$
iii) $\left(\frac{3}{10}\right)^{5} \times\left(\frac{2}{15}\right)^{5}$
$=\left[\frac{3}{10} \times \frac{2}{15}\right]^{5}$
$=\left[\frac{1}{25}\right]^{5}$
Question 6
$\frac{2^{4} \times 2 \times 3^{3} \times 7}{2^{3} \times 7^{4}}$
$=\frac{2^{4+1} \times 3^{3+6}}{2^{3} \times 3^{4}}$
$=\frac{2^{5}}{2^{3}} \times \frac{7^{9}}{7^{4}}$
$=2^{5-3} \times 9^{9-4}$
$=2^{2} \times 7^{5}$
ii) $\left(3^{2}\right)^{3} \times(-2)^{5}$
$\frac{-3^{3 \times 3} \times(-2)^{5}}{(-2)^{3}}$
$=3^{6} \times \frac{(-2)^{5}}{(-2)^{3}}$
$=3^{6} \times(-2)^{5-3}$
$=3^{6} \times(-2)^{2}$
iii) $\frac{2^{8} \times a^{5}}{4^{3} \times a^{3}}$
As $4=2^{2}$
$=\frac{2^{8} \times a^{5}}{\left(2^{2}\right)^{3} \times a^{3}}$
$=\frac{2^{8}}{2^{6}} \times \frac{a^{5}}{a^{3}}$
$=2^{8-6} \times a^{5-3}$
$=2^{2} \times a^{2}=(2 \times a)^{2}=4 a^{2}$
iv)$\frac{3 \times 7^{2} \times 11^{8}}{21 \times 11^{3}}$
As $21=7 \times 3$
$=\frac{3 \times 7^{2} \times 11^{8}}{3 \times 7 \times 11}$
$=\frac{7^{2}}{7^{1}} \times \frac{11^{8}}{11^{1}}$
$=7^{2-1} \times 11^{8-1}$
$=7^{1} \times 11^{7}$
v) $\left(2^{0}+3^{0}\right) 4$
As we rnow $a^{\circ}=1$
i.e $2^{\circ}=1 ; 3^{\circ}=1: 4^{\circ}=1$
$=(1+1) \times 1=2 \times 1=2$
vi)$3^{\circ} \times 4^{\circ} \times 5^{\circ}$
As we know $a^{\circ}=1$
$3^{0}=1 \quad ; 4^{\circ}=1 ; 5^{\circ}=1$
$=1\times1\times 1$
$=1$
Question 7
i) $\frac{25}{64}$
$\begin{array}{r|l}2&64\\ \hline 2& 32 \\ \hline 2&16\\ \hline 2& 8\\ \hline 2&4\\ \hline &2 \end{array}$
$=2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=2^{6}$
$\begin{array}{r|l}5&25\\ \hline &5\end{array}$
$=5 \times 5$
$=5^{2}$
$=\frac{5^{2}}{2^{6}}$
ii. $\frac{-125}{216}$
$\begin{array}{r|l}5&125\\ \hline 5&25\\ \hline &5\end{array}$
$=5 \times 5 \times 5$
$=5^{3}$
$\begin{array}{r|l}2&216\\ \hline 2&72 \\ \hline 2&36\\ \hline 2& 18\\ \hline 2&9\\ \hline &3 \end{array}$
$=2 \times 2 \times 2 \times 3 \times 3 \times 3$
$=2^{3} \times 3^{3}=(3 \times 2)^{3}$
$=6^{3}$
$=\frac{5^{3}}{6^{3}}$
$=\left(\frac{-5}{6}\right)^{3}$
iii $\frac{-343}{729}$
$\begin{array}{r|l}7&343\\ \hline 7&49\\ \hline &7\end{array}$
$=7 \times 7 \times 7$
$=7^{3}$
$\begin{array}{r|l}9&729\\ \hline 9&81\\ \hline &9\end{array}$
$=9 \times 9 \times 9$
$=9^{3} .$
$=\frac{-7^{3}}{9^{3}}$
$=\left(\frac{-7}{9}\right)^{3}$
Question 8
i)
$\begin{aligned} & \frac{\left(2^{5}\right)^{2} \times 7^{3}}{8^{3} \times 3} \\=& \frac{2^{5 \times 2} \times 3^{3}}{\left(2^{3}\right)^{3} \times 7} \\=& \frac{2^{10} \times 7^{3}}{2^{9} \times 7^{1}} \end{aligned}$
$=2^{10-9} \times 3^{3-1}$
$2^{1} \times 3^{2}=2 \times 49=98$
ii) $\frac{25 \times 5^{2} \times t^{8}}{10^{3} \times t^{4}}$
$10=5 \times 2 ; 25=5 \times 5=5^{2}$
$\frac{5^{2} \times 5^{2} \times t^{8}}{(5 \times 2)^{3} \times t^{4}}$
$=\frac{5^{2+2} \times t^{8}}{5^{3} \times 2^{3} \times t^{4}}$
$=\frac{5^{4}}{5^{3}} \times \frac{t^{8}}{t^{4}} \times \frac{1}{2^{3}}$
$=\frac{5^{4-3} \times t^{8-4}}{2^{3}}$
$\frac{5 \times t^{4}}{2^{3}} \cdot=\frac{5 \times t^{4}}{8}$
iii) $\frac{3^{5} \times 10^{5} \times 25}{5^{7} \times 6^{5}}$
As $10=5 \times 2 ; 25=5 \times 5 ; 6=3 \times 2$
$=\frac{3^{5} \times(5 \times 2)^{5} \times 5 \times 5}{5^{7} \times 6^{5}}$
$=\frac{3 \times 5^{5} \times 2^{5} \times 5^{2}}{5^{7} \times(3 \times 2)^{5}}$
$=\frac{3^{5} \times 2^{5} \times 5^{5+2}}{5^{7} \times 3^{5} \times 2^{5}}$
$=\frac{3^{5}}{3^{5}} \times \frac{2^{5}}{2^{5}} \times \frac{5^{7}}{5^{7}}$
$=3^{5-5} \times 2^{5-5} \times 5^{7-7}$
$=3^{0} \times 2^{\circ} \times 5^{\circ}$
$=1 \times 1 \times 1=1$
iv) $\left(\frac{-3}{5}\right)^{-3}$
As we know $a^{-n}=\frac{1}{a^{n}}$
$=\frac{1}{\left(\frac{-3}{5}\right)^{3}}$
$=\left(\frac{-5}{3}\right)^{3}$
$=\frac{(45)^{3}}{(3)^{3}}$
$=\frac{-5 \times -5 \times-5}{3 \times 3 \times 3}$
$=\frac{-125}{27}$
Question 9
i)
$\begin{aligned} &\left[\frac{-1}{2}\right]^{5} \times 2^{6} \times\left(\frac{3}{4}\right)^{3} \\=& \frac{(-1)^{5} \times 2^{6} \times 3^{3}}{(-2)^{5} \times 4^{3}} \end{aligned}$
$2^{6}=2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=64$
$3^{3}=3 \times 3 \times 3=27$
$2^{5}=2 \times 2 \times 2 \times 2 \times 2=32$
$4^{3}=4 \times 4 \times 4=64$
$=\frac{1 \times 64 \times 27}{32 \times 64}$
$=\frac{27}{32}$
ii) $\left[\left(\frac{-3}{4}\right)^{3} \div\left(-\frac{5}{2}\right)^{3}\right] \times\left(-\frac{2}{3}\right)^{4}$
$=\left(\frac{3}{42} \times \frac{2}{5}\right)^{3} \times\left(-\frac{2}{3}\right)^{4}$
$=\frac{3^{3}}{10^{3}} \times \frac{-2 \times -2 \times-2 \times -2}{3 \times 3 \times 3 \times 3}$
$\frac{27 \times 16}{1000 \times 81}$
$=\frac{2}{375}$
Question 10
i) $\left(\frac{3}{2}\right)^{-1} \div\left(-\frac{2}{5}\right)^{-1}$
$\left(\frac{3}{2}\right)^{-1}=\frac{1}{\left(\frac{3}{2}\right)}=\frac{2}{3}$
$\left(\frac{-2}{5}\right)^{-1}=\frac{1}{\left(\frac{-2}{5}\right)}=\frac{5}{-2}$
$=\frac{2}{3} \times \frac{5}{-2}$
$=\frac{-5}{3}$
ii. $\left[\left\{\left(\frac{-1}{4}\right)^{2}\right\}^{-1}\right]^{-2}$
$=\left(-\frac{1}{4}\right)^{2 \times-1 \times-2}$
$\left(\frac{-1}{4}\right)^{4}=\frac{(-1)^{4}}{4^{4}}$
$=\frac{-1 \times-1 \times -1 \times-1}{4 \times 4 \times 4\times 4}$
$=\frac{1}{256}$
Question 11
$\left(\frac{1}{3}\right)^{-2}+\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{5}\right)^{-2}-\left(\frac{1}{6}\right)^{-2}$
As we know $a^{-n}=\frac{1}{a^{n}}$
$\frac{1}{\left(\frac{1}{3}\right)^{2}}+\frac{1}{\left(\frac{1}{4}\right)^{2}}+\frac{1}{\left(\frac{1}{5}\right)^{2}}-\frac{1}{\left(\frac{1}{6}\right)^{2}}$
$=3^{2}+4^{2}+5^{2}-6^{2}$
$=9+16+25-36$
$=50-36$
$=14 .$
Question 12
i) $108 \times 192$
$\begin{array}{r|l}3&108\\ \hline 3&36 \\ \hline 3&12\\ \hline 2& 4\\ \hline &2 \end{array}$
$\begin{aligned} 108 &=3 \times 3 \times 3 \times 2 \times 2 \\ &=2^{2} \times 3^{3} \end{aligned}$
$\begin{array}{r|l}3&192\\ \hline 2&64\\ \hline 2&32\\ \hline 2& 16\\ \hline 2&8\\ \hline 2&4\\ \hline&2 \end{array}$
$\begin{aligned} 192=& 2 \times 2 \times 2 \times 2 \times 2 \times 3 \\ &=2^{5} \times 3^{1} \end{aligned}$
$\begin{aligned} 108 \times 192=& 2^{2} \times 3^{3} \times 2 \times 3^{5} \times 1 \\ &=2^{2+5} \times 3^{3+1} \\ &=2^{7} \times 3^{4} \end{aligned}$
ii)$729 \times 64$
$\begin{array}{r|l}3&729\\ \hline 3&243\\ \hline 3&81\\ \hline 3& 27\\ \hline 3&9\\ \hline&3 \end{array}$
$729=3 \times 3 \times 3 \times 3 \times 3 \times 3=3^{6}$
$\begin{array}{r|l}2&64\\ \hline 2&32\\ \hline 2&16\\ \hline 2& 8\\ \hline 2&4\\ \hline 2&2\\ \hline&1 \end{array}$
$64=2 \times 2 \times 2 \times 2 \times 2 \times 2=2^{6}$
$729 \times 64=2^{6} \times 3^{6}$
iii) $384 \times 147$
$\begin{array}{r|l}3&384\\ \hline 2&128\\ \hline 2&64\\ \hline 2&32\\ \hline 2&16\\ \hline 2&8\\ \hline2&4\\ \hline &2 \end{array}$
$\begin{aligned} 384 &=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \\ &=2^{7} \end{aligned}$
$\begin{array}{r|l}3&147\\ \hline 7&49\\ \hline7 &7\\ \hline &1\end{array}$
$\begin{aligned} 147 &=3 \times 7 \times 7 \\ &=3 \times 7^{2} \end{aligned}$
$384 \times 147=2^{7} \times 3 \times 7^{2}$
Question 13
i) $3^{3} \times 2^{2}+2^{2} \times 5^{\circ}$
$\begin{array}{r|l}2&112\\ \hline 2&56\\ \hline 2&28\\ \hline 2&14\\ \hline 2&7\\ \hline &7 \end{array}$
$=3 \times 3 \times 3 \times 2 \times 2+2 \times 2 \times 1$
$=108+4$
$=112$
$=2 \times 2 \times 2 \times 2 \times 7$
$=2^{4} \times 7^{1}$
ii)
$\begin{aligned} & 9^{2}+11^{2}-2^{2} \times 3 \times 17^{\circ} \\ & 2^{2} \times 3 \times 17^{\circ}=4 \times 3 \times 1=12 \end{aligned}$
$9^{2}=9 \times 9=81$
$11^{2}$ = $11\times 11$ = 121
$=81+121-12$
$=202-12$
$=190 .$
Question 14
i) Let the number multiplied be x
$\therefore \quad x \times 3^{4}=3^{7}$
$x=\frac{3^{3}}{3^{4}}$
$x=3^{7-4}$
$x=3^{3}=3 \times 3 \times 3$
$x=27$
ஃ 27 should be multiplied in order to get $3^{7}$
ii) Let the number multiplied be x
$(-6)^{-1} \times x=10^{-1}$
$\frac{1}{-6} \times x=\frac{1}{10}$
$x=\frac{-6}{10}$
$x=\frac{-3}{5}$
So, $\frac{-3}{5}$ Should be multiplied in order to get $10^{-1} .$
Question 15
$\left(\frac{12}{13}\right)^{4} \times\left(\frac{13}{12}\right)^{-8}=\left(\frac{12}{13}\right)^{2 x}$
$\left(\frac{12}{13}\right)^{4} \times \frac{1}{\left(\frac{13}{12}\right)^{8}}=\left(\frac{12}{13}\right)^{2 x}$
$\left(\frac{12}{13}\right)^{4} \times\left(\frac{12}{13}\right)^{8}=\left(\frac{12}{13}\right)^{2 x}$
$\left(\frac{12}{13}\right)^{4+8}=\left(\frac{12}{13}\right)^{2 x}$
$\left(\frac{12}{13}\right)^{12}=\left(\frac{12}{13}\right)^{2 x}$
Base equal , exponent should be same
2 x=12
∴ x=6
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