# Chapter-12: Exponents and Powers

Exercise-12.1

Q.1. (i)$$4^{-3}$$

(ii)$$(-5)^{-5}$$

(iii)$$(\frac{1}{2})^{-4}$$

Solution:

(i)$$4^{-2}$$

$$(4)^{-2}=\frac{1}{4^{2}}=\frac{1}{16}$$

(ii)$$(-5)^{-2}$$

$$(-5)^{-2}=\frac{1}{(-5)^{2}}=\frac{1}{25}$$

(iii)$$(\frac{1}{2})^{-4}$$

$$(\frac{1}{2})^{-2}=(\frac{2}{1})^{2}=(2)^{2}=4$$

Formula: $$a^{-m}=\frac{1}{a^{m}}$$

Q.2. (i)$$(-5)^{4}\div (-5)^{7}$$

(ii)$$(\frac{1}{4^{3}})^{2}$$

(iii)$$(-4)^{3}\times (\frac{5}{4})^{3}$$

(iv)$$(2^{-7}\div 2^{-10})\times 2^{-5}$$

(v)$$3^{-3}\times (-7)^{-3}$$

Solution:

(i)$$(-5)^{4}\div (-5)^{7}$$

$$(-5)^{4}\div (-5)^{7}=(-5)^{4-7}\\ \\ =(-5)^{-3}\\ \\ =\frac{1}{(-5)^{3}}$$

(ii)$$(\frac{1}{4^{3}})^{2}$$

$$(\frac{1}{4^{3}})^{2}=\frac{1^{2}}{(4^{3})^{2}}\\ \\ =\frac{1}{4^{3\times 2}}\\ \\ =\frac{1}{4^{6}}$$

(iii)$$(-4)^{3}\times (\frac{5}{4})^{3}$$

$$(-4)^{2}\times (\frac{5}{4})^{2}=(-4)^{2}\times \frac{5^{2}}{4^{2}}\\ \\ =[(-1)^{2}\times 4^{2}]\times \frac{5^{2}}{4^{2}}\\ \\ =4^{2-2}\times 5^{2}=4^{0}\times 5^{2}=5^{2}$$

(iv)$$(2^{-7}\div 2^{-10})\times 2^{-5}$$

$$(2^{-7}\div 2^{-10})\times 2^{-5}=2^{-7-(-10)}\times 2^{-5}\\ \\ =2^{-7+10}\times 2^{-5}\\ \\ =2^{3}\times 2^{-5}=2^{3-5}\\ \\ =2^{-2}=\frac{1}{2^{2}}$$

(v)$$3^{-3}\times (-7)^{-3}$$

$$3^{-3}\times (-7)^{-3}=\frac{1}{3^{3}}\times \frac{1}{(-7)^{3}}\\ \\ =\frac{1}{[3\times (-7)]^{3}}=\frac{1}{(-21)^{3}}$$

Formulae:

1.$$(a^{m})^{n}=a^{m\times n}$$

2.$$(\frac{a}{b})^{m}=\frac{a^{m}}{b^{m}}$$

3.$$(ab)^{m}=a^{m} b^{m}$$

4.$$a^{m}\div a^{n}=a^{m-n}$$

5.$$a^{m}\times a^{n}=a^{m+n}$$

Q.3. Find the value of:

(i)$$(3^{0}+4^{-1})\times 2^{2}$$

(ii)$$(2^{-1}\times 4^{-1})\div 2^{-2}$$

(iii)$$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}$$

(iv)$$(3^{-1}+4^{-1}+5^{-1})^{0}$$

(v)$$[(\frac{-2}{3})^{-2}]^{2}$$

Solution:

(i)$$(3^{0}+4^{-1})\times 2^{2}$$

$$(1+\frac{1}{4})\times 2^{2}\\ \\ =(\frac{4+1}{4})\times 2^{2}=\frac{5}{4}\times 2^{2}\\ \\ =5\times 2^{2-2}\\ \\ =5\times 2^{0}=5\times 1=5$$

(ii)$$(2^{-1}\times 4^{-1})\div 2^{-2}$$

$$(2^{-1}\times 4^{-1})\div 2^{-2}=(\frac{1}{2^{1}}\times \frac{1}{4^{1}})\div 2^{-2}\\ \\ =(\frac{1}{2}\times \frac{1}{2^{2}})\div 2^{-2}=\frac{1}{2^{3}}\div 2^{-2}\\ \\ =2^{-3}\div 2^{-2}=2^{-3-(-2)}=2^{-3+2}=2^{-1}\\ \\ =\frac{1}{2}$$

(iii)$$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}$$

$$(\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}=(2^{-1})^{2}+(3^{-1})^{2}+(4^{-1})^{2}\\ \\ =2^{-1\times (-2)}+3^{-1\times (-2)}+4^{-1\times (-2)}\\ \\ =2^{2}+3^{2}+4^{2}=4+9+16=29$$

(iv)$$(3^{-1}+4^{-1}+5^{-1})^{0}$$

$$(3^{-1}+4^{-1}+5^{-1})^{0}=(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{0}\\ \\ =(\frac{20+15+12}{60})^{0}=(\frac{47}{60})^{0}= 1$$

(v)$$[(\frac{-2}{3})^{-2}]^{2}$$

$$[(\frac{-2}{3})^{-2}]^{2}=(\frac{-2}{3})^{-2\times 2}\\ \\ =(\frac{-2}{3})^{-4}=(\frac{-3}{2})^{4}=\frac{81}{16}$$

Q.4. Evaluate:

(i)$$\frac{8^{-1}\times 5^{3}}{2^{-4}}$$

(ii)$$(5^{-1}\times 2^{-1})\times 6^{-1}$$

Solution:

(i)$$\frac{8^{-1}\times 5^{3}}{2^{-4}}$$

$$\frac{8^{-1}\times 5^{3}}{2^{-4}}=\frac{(2^{3})^{-1}\times 5^{3}}{2^{-4}}=\frac{2^{-3}\times 5^{3}}{2^{-4}}\\ \\ =2^{-3(-4)}\times 5^{3}=2^{-3+4}\times 5^{3}\\ \\ =2\times 125=250$$

(ii)$$(5^{-1}\times 2^{-1})\times 6^{-1}$$

$$(5^{-1}\times 2^{-1})\times 6^{-1}=(\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}\\ \\ =\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$$

Q.5. Find the value of m for which$$5^{m}\div 5^{-3}=5^{5}$$.

Solution:

$$5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}$$ $$5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}\\ \\ \Rightarrow m+3=5\\ \\ \Rightarrow m=5-3\\ \\ \Rightarrow m=2$$

Q.6. Evaluate:

(i) $$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}$$

(ii)$$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}$$

Solution:

(i) $$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}$$

$$[(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}=[(\frac{3}{1})^{1}-(\frac{4}{1})^{1}]\, (\ because a^{-m}=\frac{1}{m})\\ \\ =[3-4]=-1$$

(ii)$$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}$$

$$(\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}\\ \\ =5^{-7(-4)}\times 8^{-4(-7)}\\ \\ =5^{-7+4}\times 8^{-4+7}=5^{-3}\times 8^{3}=\frac{8^{3}}{5^{3}}\\ \\ =\frac{512}{125}$$

Q.7. Simplify:

(i)$$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)$$

(ii)$$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$$

Solution:

(i)$$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)$$

$$\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{5^{2}\times t^{-4}}{5^{-3}\times 5\times2\times t^{-8}}\\ \\ =\frac{5^{2-(-3)-1}\times t^{-4(-8)}}{2}\\ \\ =\frac{5^{2+3-1}\times t^{-4+8}}{2}=\frac{5^{4}\times t^{4}}{2}\\ \\ =\frac{625}{2}t^{4}$$

(ii)$$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$$

$$\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times (2\times 5)^{-5}\times 5^{3}}{5^{-7}\times (2\times 3)^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5}\times 5^{3}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5+3}}{5^{-7}\times 2^{-5}\times 3^{-5}}=\frac{3^{-5}\times 2^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =3^{-5-(-5)}\times 2^{-5-(-5)}\times 5^{-2-(-7)}\\ \\ =3^{-5+5}\times 2^{-5+5}\times 5^{-2+7}=3^{0}\times 2^{0}\times 5^{5}=1\times 1\times 3125\\ \\ =3125$$

Exercise-12.2

Q.1. Converts the numbers into standard form:

(i)0.0000000000085

(ii)0.00000000000942

(iii)6020000000000000

(iv)0.00000000837

(v)31860000000

Solution:

(i)0.0000000000085

$$0.0000000000085=0.0000000000085\times \frac{10^{12}}{10^{12}}=8.5\times 10^{-12}$$

(ii)0.00000000000942

$$0.00000000000942 =0.00000000000942 \times \frac{10^{12}}{10^{12}}=9.42\times 10^{-12}$$

(iii)6020000000000000

$$6020000000000000 =6020000000000000 \times \frac{10^{15}}{10^{15}}=6.02\times 10^{15}$$

(iv)0.00000000837

$$0.00000000837 =0.00000000837 \times \frac{10^{9}}{10^{9}}=8.37\times 10^{-9}$$

(v)31860000000

$$31860000000 =31860000000 \times \frac{10^{10}}{10^{10}}=3.186\times 10^{10}$$

Q.2. Convert the following numbers in usual form:

(i)$$3.02\times 10^{-6}$$

(ii)$$4.5\times 10^{4}$$

(iii)$$3\times 10^{-8}$$

(iv)$$1.0001\times 10^{9}$$

(v) $$5.8\times 10^{12}$$

(vi)$$3.61492\times 10^{6}$$

Solution:

(i)$$3.02\times 10^{-6}$$

$$3.02\times 10^{-6}=\frac{3.02}{10^{6}}=0.00000302$$

(ii)$$4.5\times 10^{4}$$

$$4.5\times 10^{4}=4.5\times 10000=45000$$

(iii) $$3\times 10^{-8}$$

$$3\times 10^{-8}=\frac{3}{10^{8}}=0.00000003$$

(iv) $$1.0001 \times 10^{9}$$

$$1.0001 \times 10^{9}=1000100000$$

(v) $$5.8\times 10^{12}$$

$$5.8\times 10^{12}=5.8\times 1000000000000=5800000000000$$

(vi) $$3.61492\times 10^{6}$$

$$3.61492\times 10^{6}=3.61492 \times 1000000 =3614920$$

Q.3. Conert the given numbers in the following statements in standard form:

(i) 1 micron is equal to $$\frac{1}{1000000}$$ m.

(ii)Charge of an electron is 0.000,000,000,000,000,000,28 coulomb.

(iii)Size of bacteria is 0.0000005 m.

(iv)Size of a plant cell is 0.00001275 m.

(v)Thickness if a thick paper is 0.07 mm.

Solution:

(i)1 micron =$$\frac{1}{1000000}=\frac{1}{10^{6}}=1\times 10^{-6}\: m$$

(ii) Charge of an electron is 0.00000000000000000028 coulombs.

$$= 0.00000000000000000028\times \frac{10^{19}}{10^{19}}=2.8\times 10^{-19}\: coulomb$$

(iii)Size of bacteria = 0.0000005 =$$\frac{5}{10000000}=\frac{5}{10^{7}}=5\times 10^{-7\: m}$$

(iv)Size of a plant cell is 0.00001275 m =$$0.00001275\times \frac{10^{5}}{10^{5}}=1.275\times 10^{-5}\: m$$

(v) Thickness of a thick paper = 0.07 mm =$$\frac{7}{100}\: mm=\frac{7}{10^{2}}=7\times 10^{-2}\: mm$$

Q.4. In a stack there are 6 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Solution:

Thickness of one book = 20 mm

Thickness of 6 books =$$20\times 6=120\: mm$$

Thickness of one paper = 0.016 mm

Thickness of 5 papers =$$0.016\times 5=0.08\: mm$$

Total thickness of a stack =120+0.08

=120.08 mm=$$120.08\times \frac{10^{2}}{10^{2}}=1.2008\times 10^{2}\: mm$$