ICSE Class 8 Maths Selina Solutions Chapter 2 Exponents (Powers)

ICSE Selina Class 8 Maths Solution of Chapter “Exponents (Powers)” is given here. The topics mentioned in the chapter explains about power which means a product of multiplying a number by itself which is usually represented with a base number and an exponent. The topic Laws of exponent with integral powers explains about three laws which need to be kept in mind while solving complex numerical problems. It includes laws of multiplication, division, double exponents, zero exponents, etc.

These detailed solutions will help students to clear all their confusion and learn about Exponents Powers in an easy and understandable way. Students should try solving the questions given in the Selina textbooks and later they can evaluate their answers by comparing with the ICSE Selina Solutions Class 8 Maths Chapter 2 “Exponents (Powers)” solutions provided here at BYJU’S website.

Download ICSE Class 8 Maths Selina Solutions PDF for Chapter 2:-Download Here

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ICSE Class 8 Maths Selina Solutions Chapter 2 Exponents (Powers) – Exercise 2 (A)

Question 1. Evaluate:

(i)

\(\begin{array}{l}\left(3^{-1} \times 9^{-1}\right) \div 3^{-2}\end{array} \)

Solution:

\(\begin{array}{l}=\left(\frac{1}{3} \times \frac{1}{9}\right) \div \frac{1}{3} \times \frac{1}{3}\end{array} \)

\(\begin{array}{l}=\frac{1}{27}\div \frac{1}{9}\end{array} \)

(Expressing the equation in fractional form)

\(\begin{array}{l}=\frac{1}{27} \times \frac{9}{1}=\frac{1}{3}\end{array} \)

(ii)

\(\begin{array}{l}\left(3^{-1} \times 4^{-1}\right) \div 6^{-1} \end{array} \)

Solution:

\(\begin{array}{l}=\left(\frac{1}{3} \times \frac{1}{4}\right) \div \frac{1}{6}\end{array} \)

\(\begin{array}{l}=\frac{1}{12} \div \frac{1}{6}\end{array} \)

(Expressing the equation in fractional form)

\(\begin{array}{l}=\frac{1}{12} \times \frac{6}{1}=\frac{1}{2}\end{array} \)

(iii)

\(\begin{array}{l}\left(2^{-1}+3^{-1}\right)^{3}\end{array} \)

Solution:

\(\begin{array}{l}=\left(\frac{1}{2}+\frac{1}{3}\right)^{3}=\left(\frac{1 \times 3}{2 \times 3}+\frac{1 \times 2}{3 \times 2}\right)^{3}\end{array} \)

\(\begin{array}{l}=\left(\frac{3+2}{6}\right)^{3}=\left(\frac{5}{6}\right)^{3}\end{array} \)

(Expressing the equation in fractional form)

\(\begin{array}{l}=\frac{5 \times 5 \times 5}{6 \times 6 \times 6}=\frac{125}{216}\end{array} \)

(iv)

\(\begin{array}{l}\left(3^{-1} \div 4^{-1}\right)^{2} \end{array} \)

Solution:

\(\begin{array}{l}=\left(\frac{1}{3} \div \frac{1}{4}\right)^{2}\end{array} \)

(Expressing the equation in fractional form)

\(\begin{array}{l}=\left(\frac{1}{3} \times \frac{4}{1}\right)^{2}=\left(\frac{4}{3}\right)^{2}\end{array} \)

(Expressing the equation in mixed fraction)

\(\begin{array}{l}=\frac{16}{9}=1 \frac{7}{9}\end{array} \)

(v)

\(\begin{array}{l}\left(2^{2}+3^{2}\right) \times\left(\frac{1}{2}\right)^{2}\end{array} \)

Solution:

\(\begin{array}{l}=(2 \times 2)+(3 \times 3) \times\left(\frac{1}{2} \times \frac{1}{2}\right) \end{array} \)

\(\begin{array}{l}=4+9 \times \frac{1}{4}=\frac{13}{4}=3 \frac{1}{4}\end{array} \)

(Simplifying the given equation)

(vi)

\(\begin{array}{l}\left(5^{2}-3^{2}\right) \times\left(\frac{2}{3}\right)^{-3}\end{array} \)

Solution:

\(\begin{array}{l}=(5 \times 5)-(3 \times 3) \times\left(\frac{3}{2}\right)^{3}\end{array} \)

\(\begin{array}{l}=25-9 \times\left(\frac{3}{2} \times \frac{3}{2} \times \frac{3}{2}\right) \end{array} \)

(Simplifying the given equation)

\(\begin{array}{l}=16 \times \frac{27}{8}=54\end{array} \)

(vii)

\(\begin{array}{l}\left[\left(\frac{1}{4}\right)^{-3}-\left(\frac{1}{3}\right)^{-3}\right]+\left(\frac{1}{6}\right)^{-3} \end{array} \)

Solution:

\(\begin{array}{l}=\left[\left(\frac{4}{1}\right)^{3}-\left(\frac{3}{1}\right)^{3}\right] \div\left(\frac{6}{1}\right)^{3}\end{array} \)

\(\begin{array}{l}=\left(\frac{4}{1} \times \frac{4}{1} \times \frac{4}{1}-\frac{3}{1} \times \frac{3}{1} \times \frac{3}{1}\right) \div\left(\frac{6}{1}\right)^{3}\end{array} \)

\(\begin{array}{l}=64-27 \times\left(\frac{1}{6} \times \frac{1}{6} \times \frac{1}{6}\right) \end{array} \)

(Simplifying the given equation)

\(\begin{array}{l}=37 \times \frac{1}{216}=\frac{37}{216}\end{array} \)

(viii)

\(\begin{array}{l}\left[\left(-\frac{3}{4}\right)^{-2}\right]^{2} \end{array} \)

Solution:

\(\begin{array}{l}\left[\left(-\frac{3}{4}\right)^{-2}\right]^{2}=\left(-\frac{3}{4}\right)^{-2 \times 2}=\left(-\frac{3}{4}\right)^{-4}\end{array} \)

\(\begin{array}{l}=\left(\frac{4}{3}\right)^{4}=\frac{4 \times 4 \times 4 \times 4}{3 \times 3 \times 3 \times 3}\end{array} \)

\(\begin{array}{l}=\frac{256}{81}=3 \frac{13}{81} \end{array} \)

(Simplifying the given equation)

(ix)

\(\begin{array}{l}\left(\left(\frac{3}{5}\right)^{-2}\right)^{-2} \end{array} \)

Solution:

\(\begin{array}{l}\left\{\left(\frac{3}{5}\right)^{-2}\right\}^{-2}=\left(\frac{3}{5}\right)^{-2 \times (-2)}=\left(\frac{3}{5}\right)^{4}\end{array} \)

\(\begin{array}{l}=\frac{3 \times 3 \times 3 \times 3}{5 \times 5 \times 5 \times 5}=\frac{81}{625} \end{array} \)

(Simplifying the given equation)

(x)

\(\begin{array}{l}\left(5^{-1} \times 3^{-1}\right)+6^{-1} \end{array} \)

Solution:

\(\begin{array}{l}=\left(\frac{1}{5} \times \frac{1}{3}\right)+\frac{1}{6}\end{array} \)

\(\begin{array}{l} =\frac{1}{15} \div \frac{1}{6}\end{array} \)

(Simplifying the given equation)

\(\begin{array}{l}=\frac{1}{15} \times \frac{6}{1}=\frac{2}{5}\end{array} \)

Question 2.

\(\begin{array}{l}1125=3^{m} \times 5^{n}\end{array} \)
; find m and n

Solution:

\(\begin{array}{l}1125=3^{2} \times 5^{3}\end{array} \)

The factors of 1125 are 3×3×5×5×5
ICSE Class 8 Maths Selina Solutions Chapter 2 Exponents (Powers) Exercise 2 (A) Question 2

therefore

\(\begin{array}{l}1125=3 \times 3 \times 5 \times 5 \times 5\end{array} \)

Now comparing,

\(\begin{array}{l}3^{2} \times 5^{3}=3^{m} \times 5^{n} \end{array} \)

therefore m=2, n=3

Question 3. Find x, if

\(\begin{array}{l} 9 \times 3^{x}=(27)^{2 x-3} \end{array} \)

Solution:

\(\begin{array}{l}9 \times 3^{x}=(27)^{2 x-3}\end{array} \)

\(\begin{array}{l}3^{2} \times 3^{x}=(3 \times 3 \times 3)^{2 x-3} \end{array} \)

(Simplifying the given equation)

\(\begin{array}{l}\Rightarrow 3^{x+2}=(3)^{3(2 x-3)} \end{array} \)

\(\begin{array}{l}\Rightarrow 3^{x+2}=(3)^{6 x-9}\end{array} \)

Since bases are same, compare them,
x+2=6x-9
6x-x=9+2

\(\begin{array}{l}\Rightarrow 5 x=11\end{array} \)

\(\begin{array}{l}\Rightarrow x=\frac{11}{2} \end{array} \)

(Shifting the terms)

\(\begin{array}{l}\Rightarrow x=2 \frac{1}{5}\end{array} \)

ICSE Class 8 Maths Selina Solutions Chapter 2 Exponents (Powers) – Exercise 2(B)

Question 1. Compute:

(i)

\(\begin{array}{l}1^{8} \times 3^{0} \times 5^{3} \times 2^{2} \end{array} \)

Solution:

\(\begin{array}{l} 1^{8} \times 3^{0} \times 5^{3} \times 2^{2} \end{array} \)

\(\begin{array}{l} {=1 \times 1 \times 5 \times 5 \times 5 \times 2 \times 2}\end{array} \)

\(\begin{array}{l}=125 \times 4 \end{array} \)
(Simplifying the given equation)

=500

(ii)

\(\begin{array}{l}\left(4^{7}\right)^{2} \times\left(4^{-3}\right)^{4} \end{array} \)

Solution:

\(\begin{array}{l} \left(4^{7}\right)^{2} \times\left(4^{-3}\right)^{4} \end{array} \)

\(\begin{array}{l}=4^{14} \times 4^{-12}\end{array} \)

\(\begin{array}{l}=4^{14-12}=4^{2} \end{array} \)
(Simplifying the given equation)

\(\begin{array}{l}=4 \times 4=16\end{array} \)

(iii)

\(\begin{array}{l}\left(2^{-9} \div 2^{-11}\right)^{3} \end{array} \)

Solution:

\(\begin{array}{l}=\left(2^{-9+11}\right)^{3}\end{array} \)

\(\begin{array}{l}=\left(2^{2}\right)^{3}=2^{6} \end{array} \)

(Simplifying the given equation)

\(\begin{array}{l}=2 \times 2 \times 2 \times 2 \times 2 \times 2=64\end{array} \)

(iv)

\(\begin{array}{l}\left(\frac{2}{3}\right)^{-4} \times\left(\frac{27}{8}\right)^{-2} \end{array} \)

Solution:

\(\begin{array}{l}\left(\frac{2}{3}\right)^{-4} \times\left(\frac{27}{8}\right)^{-2}=\left(\frac{2}{3}\right)^{-4} \times\left(\frac{3^{3}}{2^{8}}\right)^{-2}\end{array} \)

\(\begin{array}{l}=\frac{2^{-4}}{3^{-4}} \times \frac{3^{-6}}{2^{-6}}=\frac{2^{-4}}{2^{-6}} \times \frac{3^{-6}}{3^{-4}}\end{array} \)

\(\begin{array}{l}=2^{-4+6} \times \frac{1}{3^{-4+6}}=\frac{2^{2}}{3^{2}}=\frac{4}{9}\end{array} \)

(v)

\(\begin{array}{l}\left(\frac{56}{28}\right)^{0} \div\left(\frac{2}{5}\right)^{3} \times \frac{16}{25} \end{array} \)

Solution:

\(\begin{array}{l}\left(\frac{56}{28}\right)^{0} \div\left(\frac{2}{5}\right)^{3} \times \frac{16}{25}\end{array} \)

\(\begin{array}{l}=1 \div \frac{2^{3}}{5^{3}} \times \frac{2 \times 2 \times 2 \times 2}{5 \times 5}\end{array} \)

\(\begin{array}{l}\left[\left(\frac{56}{28}\right)^{0}=1\right] \end{array} \)

\(\begin{array}{l}=1 \times \frac{5^{3}}{2^{3}} \times \frac{2^{4}}{5^{2}}=5^{3-2} \times 2^{4-3}\end{array} \)

\(\begin{array}{l}=5^{1} \times 2^{1}=10\end{array} \)

(vi)

\(\begin{array}{l} (12)^{-2} \times 3^{3}\end{array} \)

Solution:

\(\begin{array}{l}=(2 \times 2 \times 3)^{-2} \times 3^{3}\end{array} \)

\(\begin{array}{l}=\left(2^{2} \times 3\right)^{-2} \times 3^{3}\end{array} \)

\(\begin{array}{l}=2^{-2 \times 2} \times 3^{-2} \times 3^{3}\end{array} \)

\(\begin{array}{l}=2^{-4} \times 3^{-2+3} \times 3^{3}\end{array} \)

\(\begin{array}{l}=2^{-4} \times 3^{1}\end{array} \)

\(\begin{array}{l}=\frac{3}{2^{4}}=\frac{3}{2 \times 2 \times 2 \times 2}=\frac{3}{16}\end{array} \)

(vii)

\(\begin{array}{l} (-5)^{4} \times(-5)^{6} \div(-5)^{9}\end{array} \)

Solution:

\(\begin{array}{l}=(-5)^{4} \times(-5)^{6} \times \frac{1}{(-5)^{9}}\end{array} \)

\(\begin{array}{l}=(-5)^{4+6-9}\end{array} \)

\(\begin{array}{l}=(-5)^{1}=-5\end{array} \)

(viii)

\(\begin{array}{l}\left(-\frac{1}{3}\right)^{4} \div\left(-\frac{1}{3}\right)^{8} \times\left(-\frac{1}{3}\right)^{5}\end{array} \)

Solution:

\(\begin{array}{l}=\left(-\frac{1}{3}\right)^{4} \times \frac{1}{\left(-\frac{1}{3}\right)^{8}} \times\left(-\frac{1}{3}\right)^{5}\end{array} \)

\(\begin{array}{l}=\left(-\frac{1}{3}\right)^{4+5-8}=\left(-\frac{1}{3}\right)^{9-8}\end{array} \)

\(\begin{array}{l}=-\frac{1}{3}\end{array} \)

(ix)

\(\begin{array}{l}9^{0} \times 4^{-1} \div 2^{-4}\end{array} \)

Solution:

\(\begin{array}{l}9^{0} \times 4^{-1} \div 2^{-4}=1 \times \frac{1}{4^{1}} \times \frac{1}{2^{-4}}\end{array} \)

\(\begin{array}{l}=1 \times \frac{1}{4} \times 2^{4}=1 \times \frac{1}{2^{2}} \times 2^{4}\end{array} \)

\(\begin{array}{l}=2^{4-2}=2^{2}=4\end{array} \)

(x)

\(\begin{array}{l} (625)^{-\frac{3}{4}} \end{array} \)

Solution:

\(\begin{array}{l} (625)^{-\frac{3}{4}}=(5 \times 5 \times 5 \times 5)^{-\frac{3}{4}}\end{array} \)

\(\begin{array}{l}=\left(5^{4}\right)^{-\frac{3}{4}}=5^{4 \times-\frac{3}{4}}\end{array} \)

\(\begin{array}{l}=5^{-3}=\frac{1}{5^{3}}\end{array} \)

\(\begin{array}{l}=\frac{1}{5 \times 5 \times 5}\end{array} \)

\(\begin{array}{l}=\frac{1}{125}\end{array} \)

(xi)

\(\begin{array}{l}\left(\frac{27}{64}\right)^{-\frac{2}{3}} \end{array} \)

Solution:

\(\begin{array}{l}\left(\frac{27}{64}\right)^{-\frac{2}{3}}=\left[\frac{\left(3^{3}\right)}{\left(4^{3}\right)}\right]^{-\frac{2}{3}}\end{array} \)

\(\begin{array}{l}=\frac{3^{3 \times-\frac{2}{3}}}{4^{3 \times-\frac{2}{3}}}=\frac{3^{-2}}{4^{-2}}\end{array} \)

\(\begin{array}{l}=\frac{4^{2}}{3^{2}}=\frac{4 \times 4}{3 \times 3}=\frac{16}{9}=1 \frac{7}{9}\end{array} \)

(xii)

\(\begin{array}{l}\left(\frac{1}{32}\right)^{-\frac{2}{5}}\end{array} \)

Solution:

\(\begin{array}{l}\left(\frac{1}{32}\right)^{-\frac{2}{5}}=\left(\frac{1}{2 \times 2 \times 2 \times 2 \times 2}\right)^{\frac{2}{5}}\end{array} \)

\(\begin{array}{l}=\left(\frac{1}{2^{5}}\right)^{-\frac{2}{5}}=\frac{1}{2^{5 \times -\frac{2}{5}}}\end{array} \)

\(\begin{array}{l}=\frac{1}{2^{-2}}=2^{2}=4\end{array} \)

(xiii)

\(\begin{array}{l} (125)^{-\frac{2}{3}} \div(8)^{\frac{2}{3}}\end{array} \)

Solution:

\(\begin{array}{l} (125)^{-\frac{2}{8}} \div(8)^{\frac{2}{3}}=\left(5^{3}\right)^{-\frac{2}{3}} \div\left(2^{3}\right)^{\frac{2}{3}}\end{array} \)

\(\begin{array}{l}=5^{-\frac{2}{3} \times 3} \div 2^{3 \times \frac{2}{3}}\end{array} \)

\(\begin{array}{l}=5^{-2} \div 2^{2}=\frac{1}{5^{2}} \times \frac{1}{2^{2}}\end{array} \)

\(\begin{array}{l}=\frac{1}{25} \times \frac{1}{4}=\frac{1}{100}\end{array} \)

(xiv)

\(\begin{array}{l} (243)^{\frac{2}{5}} \div(32)^{-\frac{2}{5}} \end{array} \)

Solution:

\(\begin{array}{l} =(3 \times 3 \times 3 \times 3 \times 3)^{\frac{2}{5}} \div(2 \times 2 \times 2 \times 2 \times 2)^{-\frac{2}{5}} \end{array} \)

\(\begin{array}{l}=\left(3^{5}\right)^{\frac{2}{5}} \div\left(2^{5}\right)^{-\frac{2}{5}}\end{array} \)

\(\begin{array}{l}=3^{5 \times \frac{2}{5}} \div 2^{-\frac{2}{5} \times 5}=3^{2} \div 2^{-2}\end{array} \)

\(\begin{array}{l}=3^{2} \times \frac{1}{2^{-2}}=3^{2} \times 2^{+2}\end{array} \)

\(\begin{array}{l}=3 \times 3 \times 2 \times 2=36\end{array} \)

(xv)

\(\begin{array}{l} (-3)^{4}-(\sqrt[4]{3})^{0} \times(-2)^{5} \div(64)^{\frac{2}{3}}\end{array} \)

Solution:

\(\begin{array}{l}=(-3 \times-3 \times-3 \times-3) -1 \times-2 \times-2 \times-2 \times-2 \times-2 \div\left(2^{6}\right)^{\frac{2}{3}}\end{array} \)

Note:

\(\begin{array}{l} (\sqrt[4]{3})^{0}=1\end{array} \)

\(\begin{array}{l}=3^{4}+2^{5} \div 2^{6 \times \frac{2}{3}}\end{array} \)

\(\begin{array}{l}=3^{4}+2^{5} \div 2^{4}=3^{4}+\frac{2^{5}}{2^{4}}\end{array} \)

\(\begin{array}{l}=3^{4}+2^{5-4}=3^{4}+2=3 \times 3 \times 3 \times 3+2\end{array} \)

=81+2=83

(xvi)

\(\begin{array}{l} (27)^{\frac{2}{3}} \div\left(\frac{81}{16}\right)^{-\frac{1}{4}}\end{array} \)

Solution:

\(\begin{array}{l} (27)^{\frac{2}{3}} \div\left(\frac{81}{16}\right)^{-\frac{1}{4}}=\left(3^{3}\right)^{\frac{2}{3}} \div\left(\frac{3^{4}}{2^{4}}\right)^{-\frac{1}{4}} \end{array} \)

\(\begin{array}{l} =3^{3 \times \frac{2}{3}} \div \frac{3^{-\frac{1}{4} \times 4}}{2^{-\frac{1}{4} \times 4}}=3^{2} \div \frac{3^{-1}}{2^{-1}} \end{array} \)

\(\begin{array}{l}=3^{2} \times \frac{2^{-1}}{3^{-1}}\end{array} \)

\(\begin{array}{l}=3^{2+1} \times 2^{-1}=3^{3} \times \frac{1}{2^{+1}}\end{array} \)

\(\begin{array}{l}=\frac{3 \times 3 \times 3}{2}=\frac{27}{2}=13 \frac{1}{2}\end{array} \)

Question 2. Simplify:

(i)

\(\begin{array}{l} 8^{\frac{4}{3}}+25^{\frac{3}{2}}-\left(\frac{1}{27}\right)^{-\frac{2}{3}} \end{array} \)

Solution:

\(\begin{array}{l}=\left(2^{3}\right)^{\frac{4}{3}}+\left(5^{2}\right)^{\frac{3}{2}}-\left(\frac{1}{3^{3}}\right)^{-\frac{2}{3}}\end{array} \)

\(\begin{array}{l}=2^{3 \times \frac{4}{3}}+5^{2 \times \frac{3}{2}}-\frac{1}{3^{3}\times\left(\frac{-2}{3}\right)} \end{array} \)

\(\begin{array}{l}=2^{4}+5^{3}-\frac{1}{3^{-2}}\end{array} \)

\(\begin{array}{l}=16+125-3^{2}\end{array} \)

=141-9=132

(ii)

\(\begin{array}{l} (64)^{-2} ]^{-3} \div\left[\left\{(-8)^{2}\right\}^{3}\right]^{2} \end{array} \)

Solution:

\(\begin{array}{l}=\left(2^{6}\right)^{-2 \times-3} \div(-8)^{2 \times 3 \times 2}\end{array} \)

\(\begin{array}{l}=2^{6 \times(6)} \div(-8)^{12}\end{array} \)

\(\begin{array}{l}=2^{+36} \div(-8)^{12} \end{array} \)

\(\begin{array}{l}=2^{+36} \div\left[(-2)^{3}\right]^{12}=2^{36} \div(-2)^{36}\end{array} \)

\(\begin{array}{l}=\frac{2^{36}}{(-2)^{36}}=\frac{2^{36}}{2^{36}}\end{array} \)
\(\begin{array}{l} (\ 36 \text { is even })\end{array} \)

\(\begin{array}{l}=2^{36-36}=2^{0}=1 \end{array} \)
(therefore
\(\begin{array}{l}a^{0}=1\end{array} \)
)

(iii)

\(\begin{array}{l}\left(2^{-3}-2^{-4}\right)\left(2^{-3}+2^{-4}\right) \end{array} \)

Solution:

\(\begin{array}{l}=\left(2^{-3}\right)^{2}-\left(2^{-4}\right)^{2}\end{array} \)

\(\begin{array}{l}\left\{(a-b)(a+b)=a^{2}-b^{2}\right\}\end{array} \)

\(\begin{array}{l}=2^{-6}-2^{-8}=\frac{1}{2^{6}}-\frac{1}{2^{8}}\end{array} \)

\(\begin{array}{l}=\frac{1}{64}-\frac{1}{256}\end{array} \)

\(\begin{array}{l}=\frac{4-1}{256}=\frac{3}{256}\end{array} \)

Question 3. Evaluate:

(i)

\(\begin{array}{l} (-5)^{0} \end{array} \)

Solution:

\(\begin{array}{l} (-5)^{0}=1\left( a^{0}=1\right) \end{array} \)

(ii)

\(\begin{array}{l}8^{0}+4^{0}+2^{0} \end{array} \)

Solution:

\(\begin{array}{l}8^{0}+4^{0}+2^{0}=1+1+1=3\end{array} \)
\(\begin{array}{l} ( a^{0}=1) \end{array} \)

(iii)

\(\begin{array}{l} (8+4+2)^{0} \end{array} \)

Solution:

\(\begin{array}{l} (8+4+2)^{0}=(14)^{0}=1 \end{array} \)
\(\begin{array}{l}( a^{0}=1) \end{array} \)

(iv)

\(\begin{array}{l}4x^{0} \end{array} \)

Solution:

\(\begin{array}{l}4x^{0}=4 \times 1=4\end{array} \)

(v)

\(\begin{array}{l} (4x)^{0} \end{array} \)

Solution:

\(\begin{array}{l} (4x)^{0}=1\end{array} \)

(vi)

\(\begin{array}{l}\left[\left(10^{3}\right)^{0}\right]^{5}\end{array} \)

Solution:

\(\begin{array}{l}\left[\left(10^{3}\right)^{0}\right]^{5}=10^{3 \times 0 \times 5}=10^{0}=1\end{array} \)

(vii)

\(\begin{array}{l}\left(7x^{0}\right)^{2} \end{array} \)

Solution:

\(\begin{array}{l}\left(7x^{0}\right)^{2}=7^{2} \times x^{0 \times 2}=49 \times 1=49\end{array} \)

(viii)

\(\begin{array}{l}9^{0}+9^{-1}-9^{-2}+9^{\frac{1}{2}}-9^{-\frac{1}{2}} \end{array} \)

Solution:

\(\begin{array}{l}9^{0}+9^{-1}-9^{-2}+\frac{1}{9^{\frac{1}{2}}}-9^{-\frac{1}{2}}\end{array} \)

\(\begin{array}{l} =1+\frac{1}{9}-\frac{1}{9^{2}}+\left(3^{2}\right)^{\frac{1}{2}}-\left(3^{2}\right)^{-\frac{1}{2}} \end{array} \)

\(\begin{array}{l}=1+\frac{1}{9}-\frac{1}{81}+3^{2 \times \frac{1}{2}}-3^{2 \times\left(-\frac{1}{2}\right)} \end{array} \)

\(\begin{array}{l}=1+\frac{1}{9}-\frac{1}{81}+3-3^{-1}\end{array} \)

\(\begin{array}{l}=1+\frac{1}{9}-\frac{1}{81}+\frac{3}{1}-\frac{1}{3}\end{array} \)

\(\begin{array}{l}=\frac{81+9-1+243-27}{81}=\frac{333-28}{81}\end{array} \)

\(\begin{array}{l}=\frac{305}{81}=3 \frac{62}{81}\end{array} \)

Question 4. Simplify:

(i)

\(\begin{array}{l}\frac{a^{5} b^{2}}{a^{2} b^{-3}} \end{array} \)

Solution:

\(\begin{array}{l}\frac{a^{5} b^{2}}{a^{2} b^{-3}}=a^{5-2} \cdot b^{2+3}=a^{3} b^{5}\end{array} \)

(ii)

\(\begin{array}{l}15 y^{8} \div 3 y^{3} \end{array} \)

Solution:

\(\begin{array}{l}15 y^{8} \div 3 y^{3}=\frac{15 y^{8}}{3 y^{3}}\end{array} \)

\(\begin{array}{l}=5 y^{\{8-3\}}\end{array} \)

\(\begin{array}{l}=5 y^{5}\end{array} \)

(iii)

\(\begin{array}{l}x^{10} y^{6} \div x^{3} y^{-2} \end{array} \)

Solution:

\(\begin{array}{l}x^{10} y^{6} \div x^{3} y^{-2}=\frac{x^{10} y^{6}}{x^{3} y^{-2}}\end{array} \)

\(\begin{array}{l}=x^{10-3} \cdot y^{6+2}\end{array} \)

\(\begin{array}{l}=x^{7} y^{8}\end{array} \)

(iv)

\(\begin{array}{l}5z^{16} \div 15z^{-11}\end{array} \)

Solution:

\(\begin{array}{l}5z^{16} \div 15z^{-11}=\frac{5z^{16}}{15z^{-11}}\end{array} \)

\(\begin{array}{l}=\frac{5}{15} z^{16+11}\end{array} \)

\(\begin{array}{l}=\frac{1}{3} z^{27}\end{array} \)

(v)

\(\begin{array}{l}\left(36x^{2}\right)^{\frac{1}{2}} \end{array} \)

Solution:

\(\begin{array}{l}\left(36x^{2}\right)^{\frac{1}{2}}=(36)^{\frac{1}{2}} \cdot x^{2 \times \frac{1}{2}}\end{array} \)

\(\begin{array}{l}=(6 \times 6)^{\frac{1}{2}} \cdot x=\left(6^{2}\right)^{\frac{1}{2}} \cdot x\end{array} \)

= 6x

(vi)

\(\begin{array}{l}\left(125x^{-3}\right)^{\frac{1}{3}}\end{array} \)

Solution:

\(\begin{array}{l} \left(125 x^{-3}\right)^{\frac{1}{3}}=(125)^{\frac{1}{3}} x^{-3 \times \frac{1}{3}} \end{array} \)

\(\begin{array}{l}=(5 \times 5 \times 5)^{\frac{1}{3}} x^{-1}\end{array} \)

\(\begin{array}{l}\left(5^{3}\right)^{\frac{1}{3}} \cdot x^{-1}=5 x^{-1}\end{array} \)

\(\begin{array}{l}=\frac{5}{x}=5 x^{-1}\end{array} \)

(vii)

\(\begin{array}{l}\left(2x^{2} y^{-3}\right)^{-2}\end{array} \)

Solution:

\(\begin{array}{l}\left(2x^{2} y^{-3}\right)^{-2}=2^{-2} x^{2 \times-2} \cdot y^{-3x-2}\end{array} \)

\(\begin{array}{l}=\frac{1}{2^{2}} x^{-4} \cdot y^{6}\end{array} \)

\(\begin{array}{l}=\frac{1}{4} \times \frac{y^{6}}{x^{4}}\end{array} \)

\(\begin{array}{l}=\frac{y^{6}}{4x^{4}}=\frac{1}{4} \cdot y^{6} x^{-4}\end{array} \)

(viii)

\(\begin{array}{l}\left(27 x^{-3} y^{6}\right)^{\frac{2}{3}} \end{array} \)

Solution:

\(\begin{array}{l}\left(27 x^{-3} y^{6}\right)^{\frac{2}{3}}=(27)^{\frac{2}{3}} \cdot x^{-3 \times \frac{2}{3}} \cdot y^{6 \times \frac{2}{3}}\end{array} \)

\(\begin{array}{l}=(3 \times 3 \times 3)^{\frac{2}{3}} x^{-2} \cdot y^{4}\end{array} \)

\(\begin{array}{l}=\left[(3 \times 3 \times 3)^{\frac{1}{3}}\right]^{2} x^{-2} \cdot y^{4}\end{array} \)

\(\begin{array}{l}=3^{2} x^{-2} y^{4}\end{array} \)

\(\begin{array}{l}=9x^{-2} y^{4}\end{array} \)

\(\begin{array}{l}=\frac{9 y^{4}}{x^{2}}=9x^{-2} y^{4}\end{array} \)

(ix)

\(\begin{array}{l}\left(-2 x^{\frac{2}{3}} y^{-\frac{3}{2}}\right)^{6} \end{array} \)

Solution:

\(\begin{array}{l} =(-2)^{6} x^{\frac{2}{3} \times 6} y^{-\frac{3}{2} \times 6} \end{array} \)

\(\begin{array}{l}=64x^{4} y^{-9}\end{array} \)

\(\begin{array}{l}=\frac{64x^{4}}{y^{9}}\end{array} \)

\(\begin{array}{l}=64 x^{4} y^{-9}\end{array} \)

Question 5. Simplify:

\(\begin{array}{l}\left(x^{a+b}\right)^{a-b} \cdot\left(x^{b+c}\right)^{b-c} \cdot\left(x^{c+a}\right)^{c-a}\end{array} \)

Solution:

\(\begin{array}{l}\left(x^{a+b}\right)^{a-b} \cdot\left(x^{b+c}\right)^{b-c} \cdot\left(x^{c+a}\right)^{c-a}\end{array} \)

\(\begin{array}{l}=x^{(a+b)(a-b)} x^{(b+c)(b-c)} x^{(c+a)(c-a)} \end{array} \)

\(\begin{array}{l}=x^{a^{2}-b^{2}} x^{b^{2}-c^{2}} x^{c^{2}-a^{2}}\end{array} \)

\(\begin{array}{l}=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\end{array} \)

\(\begin{array}{l}=x^{0}\end{array} \)

=1

Question 6.
Simplify:

(i)

\(\begin{array}{l}\sqrt[5]{x^{20} y^{-10} z^{5}}\div \frac{x^{3}}{y^{3}} \end{array} \)

Solution:

\(\begin{array}{l}\sqrt[5]{x^{20} y^{-10} z^{5}} \div \frac{x^{3}}{y^{3}}\end{array} \)

\(\begin{array}{l}=\left(x^{20} y^{-10} z^{5}\right)^{\frac{1}{5}} \div \frac{x^{3}}{y^{3}}\end{array} \)

\(\begin{array}{l}x^{20 \times \frac{1}{5}} \cdot y^{-10 \times \frac{1}{5}} \cdot z^{5 \times \frac{1}{5}} \div \frac{x^{3}}{y^{3}}\end{array} \)

\(\begin{array}{l}=x^{4} \cdot y^{-2} \cdot z^{1} \times \frac{y^{3}}{x^{3}}\end{array} \)

\(\begin{array}{l}=x^{4-3} \cdot y^{-2+3} \cdot z^{1} \end{array} \)

= xyz

(ii)

\(\begin{array}{l}\left(\frac{256 a^{16}}{81b^{4}}\right)^{\frac{-3}{4}}\end{array} \)

Solution:

\(\begin{array}{l}\left[\frac{256 a^{16}}{81 b^{4}}\right]^{-\frac{3}{4}}=\left[\frac{4^{4} a^{16}}{3^{4} b^{4}}\right]^{\frac{-3}{4}}\end{array} \)

Where

\(\begin{array}{l}256=4 \times 4 \times 4 \times 4=4^{4}\end{array} \)

\(\begin{array}{l}81=3 \times 3 \times 3 \times 3=3^{4}\end{array} \)

\(\begin{array}{l}=\frac{4^{4 \times \frac{-3}{4}} \cdot a^{16 \times \frac{-3}{4}}}{3^{4 \times \frac{-3}{4}} b^{4 \times \frac{-3}{4}}}\end{array} \)

\(\begin{array}{l}=\frac{4^{-3} \cdot a^{-12}}{3^{-3} \cdot b^{-8}} \end{array} \)

\(\begin{array}{l}=\frac{3^{3} b^{3}}{4^{3} a^{12}}\end{array} \)

\(\begin{array}{l}=\frac{27 b^{3}}{64 a^{12}}\end{array} \)

\(\begin{array}{l}=\frac{27}{64} \cdot a^{-12} b^{3}\end{array} \)

Question 7

{i}

\(\begin{array}{l}\left(a^{-2}\right)^{-2} \cdot(ab)^{-3}\end{array} \)

Solution:

\(\begin{array}{l}\left(a^{-2}\right)^{-2} \cdot(ab)^{-3}\end{array} \)

\(\begin{array}{l}=\left(a^{-2 \times -2} \cdot b^{-2}\right) \cdot\left(a^{-3} \cdot b^{-3}\right) \end{array} \)

\(\begin{array}{l}=a^{+4} \cdot b^{-2} \cdot a^{-3} \cdot b^{-3}\end{array} \)

\(\begin{array}{l}=a^{4-3} \cdot b^{-2-3} \end{array} \)

\(\begin{array}{l}=a b^{-5}\end{array} \)

\(\begin{array}{l}=\frac{a}{b^{5}}\end{array} \)

(ii)

\(\begin{array}{l}\left(x^{n} y^{-m}\right)^{4} \times\left(x^{3} y^{-2}\right)^{-n}\end{array} \)

Solution:

\(\begin{array}{l}\left(x^{n} y^{-m}\right)^{4} \times\left(x^{3} y^{-2}\right)^{-n} =x^{4n} y^{-4m} \times x^{-3n} y^{2n} \end{array} \)

\(\begin{array}{l}=x^{4n-3n} \cdot y^{-4m+2n}\end{array} \)

\(\begin{array}{l}=x^{n} y^{-4m+2n}\end{array} \)

(iii)

\(\begin{array}{l}\left(\frac{125 a^{-3}}{y^{6}}\right)^{\frac{-1}{3}}\end{array} \)

Solution:

\(\begin{array}{l}\left[\frac{125 a^{-3}}{y^{6}}\right]^{ \frac{-1}{3}}=\left[\frac{5^{3} a^{-3}}{y^{6}}\right]^{ \frac{-1}{3}}\end{array} \)

Where

\(\begin{array}{l}125=5 \times 5 \times 5=5^{3}\end{array} \)

\(\begin{array}{l}=\frac{5^{3 \times \frac{-1}{3}} \cdot a^{-3 \times \frac{-1}{3}}}{y^{6 \times \frac{-1}{3}}}\end{array} \)

\(\begin{array}{l}=\frac{5^{-1} \cdot a^{1}}{y^{-2}}\end{array} \)

\(\begin{array}{l} =\frac{a \cdot y^{2}}{5} \end{array} \)

(iv)

\(\begin{array}{l}\left(\frac{32 x^{-5}}{243 y^{-5}}\right)^{\frac{-1}{5}}\end{array} \)

Solution:

\(\begin{array}{l}\left[\frac{32 x^{-5}}{243 y^{-5}}\right]^{\frac{-1}{5}}=\left[\frac{2^{5} x^{-5}}{3^{5} y^{-5}}\right]^{\frac{-1}{5}}\end{array} \)

Where

\(\begin{array}{l}32=2 \times 2 \times 2 \times 2 \times 2=2^{5} \end{array} \)

\(\begin{array}{l}243=3 \times 3 \times 3 \times 3 \times 3=3^{5}\end{array} \)

\(\begin{array}{l}=\frac{2^{5 \times \frac{-1}{5}} \cdot x^{-5 \times \frac{-1}{5}}}{3^{5 \times \frac{-1}{5}} y^{-5 \times \frac{-1}{5}}}\end{array} \)

\(\begin{array}{l}=\frac{2^{-1} x^{+1}}{3^{-1} y^{+1}}\end{array} \)

\(\begin{array}{l}=\frac{3x}{2y}\end{array} \)

(v)

\(\begin{array}{l}\left(a^{-2} b\right)^{\frac{1}{2}} \times\left(ab^{-3}\right)^{\frac{1}{3}}\end{array} \)

Solution:

\(\begin{array}{l}\left(a^{-2} b\right)^{\frac{1}{2}} \times\left(ab^{-3}\right)^{\frac{1}{3}}\end{array} \)

\(\begin{array}{l}=\left(a^{-2 \times \frac{1}{2}} \cdot b^{\frac {1}{2}}\right) \times\left(a^{\frac{1}{3}} b^{-3 \times \frac{1}{3}}\right) \end{array} \)

\(\begin{array}{l}=a^{-1} b^{\frac{1}{2}} \times a^{\frac{1}{3}} b^{-1}\end{array} \)

\(\begin{array}{l}=a^{-1+\frac{1}{3}} b^{\frac{1}{2}-1}\end{array} \)

\(\begin{array}{l}=a^{-\frac{2}{3}} b^{-\frac{1}{2}}\end{array} \)

\(\begin{array}{l}=\frac{1}{a^{\frac{2}{3}} b^{\frac{1}{2}}}\end{array} \)

(vi)

\(\begin{array}{l} (xy)^{m-n} \cdot(yz)^{n-l} \cdot(zx)^{I-m}\end{array} \)

Solution:

\(\begin{array}{l} (xy)^{m-n} \cdot (yz)^{n-l} \cdot (x z)^{l-m}\end{array} \)

\(\begin{array}{l}=x^{m-n} \cdot y^{m-n} \cdot y^{n-l} \cdot z^{n-l} x^{l-m} \cdot z^{l-m}\end{array} \)

\(\begin{array}{l}=x^{m-n+l-m} \cdot y^{m-n+n-l} \cdot z^{n-l+l-m}\end{array} \)

\(\begin{array}{l}=x^{l-n} \cdot y^{m-l} \cdot z^{n-m}\end{array} \)

Question 8.

Show that:

\(\begin{array}{l}\left(\frac{x^{a}}{x^{-b}}\right)^{a-b} \cdot \left(\frac{x^{b}}{x^{-c}}\right)^{b-c} \cdot \left(\frac{x^{c}}{x^{-a}}\right)^{c-a}=1\end{array} \)

Solution:

L.H.S.

\(\begin{array}{l}=\left(\frac{x^{a}}{x^{-b}}\right)^{a-b} \cdot \left(\frac{x^{b}}{x^{-c}}\right)^{b-c} \cdot \left(\frac{x^{c}}{x^{-a}}\right)^{c-a}\end{array} \)

\(\begin{array}{l}=\left(x^{a+b}\right)^{a-b} \cdot\left(x^{b+c}\right)^{b-c} \cdot\left(x^{c+a}\right)^{c-a}\end{array} \)

\(\begin{array}{l}=x^{(a+b)(a-b)} x^{(b+c)(b-c)} x^{(c+a)(c-a)} \end{array} \)

\(\begin{array}{l}=x^{a^{2}-b^{2}} x^{b^{2}-c^{2}} x^{c^{2}-a^{2}}\end{array} \)

\(\begin{array}{l}=x^{a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}}\end{array} \)

\(\begin{array}{l}=x^{0}\end{array} \)

=1 = R.H.S

Question 9.

Evaluate:

\(\begin{array}{l}\frac{x^{5+n}\left(x^{2}\right)^{3n+1}}{x^{7n-2}}\end{array} \)

Solution:

\(\begin{array}{l}\frac{x^{5+n} \left(x^{2}\right)^{3n+1}}{x^{7n-2}}\end{array} \)

\(\begin{array}{l}=\frac{x^{5+n} \times x^{2(3n+1)}}{x^{7n-2}}\end{array} \)

\(\begin{array}{l}=\frac{x^{5+n} \times x^{6n+2}}{x^{7n-2}}\end{array} \)

\(\begin{array}{l}=x^{5+n+6n+2-7n+2}\end{array} \)

\(\begin{array}{l}=x^{9}\end{array} \)

Question 10. Evaluate:

\(\begin{array}{l}\frac{a^{2n+1} \times a^{(2n+1)(2n-1)}}{a^{n(4n-1) \times\left(a^{2}\right)^{2n+3}}}\end{array} \)

Solution:

\(\begin{array}{l}\frac{a^{2n+1} \times a^{(2n+1)(2n-1)}}{a^{n(4n-1) \times\left(a^{2}\right)^{2n+3}}}\end{array} \)

\(\begin{array}{l}=\frac{a^{2n+1} \times a^{(2n)^{2}-(1)^{2}}}{a^{4n^{2}-n} \times a^{2(2n+3)}} \end{array} \)

\(\begin{array}{l}=\frac{a^{2n+1} \times a^{4n^{2}-1}}{a^{4n^{2}-n} \times a^{4n+6}}\end{array} \)

\(\begin{array}{l}=a^{2n+1+4n^{2}-1-4n^{2}+n-4n-6}\end{array} \)

\(\begin{array}{l}=a^{-n-6}\end{array} \)

\(\begin{array}{l}=a^{-(n+6)} \end{array} \)

\(\begin{array}{l}=\frac{1}{a^{n+6}}\end{array} \)

Question 11.

\(\begin{array}{l} (m+n)^{-1}\left(m^{-1}+n^{-1}\right)=(m n)^{-1} \end{array} \)

Solution:

L.H.S.

\(\begin{array}{l} =(m+n)^{-1}\left(m^{-1}+n^{-1}\right) \end{array} \)

\(\begin{array}{l}=\frac{1}{m+n}\left(\frac{1}{m}+\frac{1}{n}\right)=\frac{1}{m+n} \cdot \frac{n+m}{m n}=\frac{1}{m n}\end{array} \)

\(\begin{array}{l}=(m n)^{-1}\end{array} \)

=R.H.S.
Hence proved.

Question 12. Prove that:

(i)

\(\begin{array}{l}\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{a b}}\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{b c}}\left(\frac{x^{c}}{x^{a}}\right)^{\frac{1}{c a}}=1\end{array} \)

Solution:

\(\begin{array}{l}\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{a b}}\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{b c}}\left(\frac{x^{c}}{x^{a}}\right)^{\frac{1}{c a}}=1\end{array} \)

L.H.S

\(\begin{array}{l}=\left(\frac{x^{a}}{x^{b}}\right)^{\frac{1}{a b}}\left(\frac{x^{b}}{x^{c}}\right)^{\frac{1}{b c}}\left(\frac{x^{c}}{x^{a}}\right)^{\frac{1}{c a}}\end{array} \)

\(\begin{array}{l}=\left(x^{a-b}\right)^{\frac{1}{b b}}\left(x^{b-c}\right)^{\frac{1}{b_{c}}}\left(x^{c-a}\right)^{\frac{1}{m}}\end{array} \)

\(\begin{array}{l}=x^{\frac{a-b}{a b}} \cdot x^{\frac{b-c}{b c}} \cdot x^{\frac{c-a}{c a}}\end{array} \)

\(\begin{array}{l}=x^{\frac{a-b}{a b}+\frac{b-c}{b c}+\frac{c-a}{c a}}\end{array} \)

\(\begin{array}{l}=x^{\frac{a c-b c+a b-a c+b c-a b}{a b c}}\end{array} \)

\(\begin{array}{l}=x^{0}=1 \end{array} \)
=R.H.S

(ii)

\(\begin{array}{l}\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1\end{array} \)

Solution:

\(\begin{array}{l}\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1\end{array} \)

L.H.S.

\(\begin{array}{l}=\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}\end{array} \)

\(\begin{array}{l}=\frac{1}{x^{a-a}+x^{a-b}}+\frac{1}{x^{b-b}+x^{b-a}}\end{array} \)

\(\begin{array}{l}=\frac{1}{x^{a} x^{-a}+x^{a} x^{-b}}+\frac{1}{x^{b} x^{-b}+ x^{b} x^{-a}}\end{array} \)

\(\begin{array}{l}=\frac{1}{x^{a}\left(x^{-a}+x^{-b}\right)}+\frac{1}{x^{b}\left(x^{-b}+x^{-a}\right)} \end{array} \)

\(\begin{array}{l}=\frac{1}{\left(x^{-a}+x^{-b}\right)}\left[\frac{1}{x^{a}}+\frac{1}{x^{b}}\right] \end{array} \)

\(\begin{array}{l}=\frac{1}{x^{-a}+x^{-b}}\left[x^{-a}+x^{-b}\right] \end{array} \)

= 1 = R.H.S

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