Chapter 13: Exponents and Powers

 

 

Find the value of:

(i)             $2^6$

(ii)          $9^3$ 

(iii)      ${11}^2$ 

(iv)       $5^4$

(i)             $2^6=2\times 2\times 2\times 2\times 2\times 2=64$

(ii)          $9^3=9\times 9\times 9=729$

(iii)      ${11}^2=11\times 11=121$

(iv)       $5^4=5\times 5\times 5\times 5=625$

Express the following in exponential form:

(i)             $6\times 6\times 6\times 6$

(ii)          $t\times t$

(iii)      $b\times b\times b\times b$

(iv)       $5\times 5\times 7\times 7\times 7$

(v)          $2\times 2\times a\times a$

(vi)       $a\times a\times a\times c\times c\times c\times c\times d$

(i)             $6\times 6\times 6\times 6=6^4$

(ii)          $t\times t=t^2$

(iii)      $b\times b\times b\times b=b^4$

(iv)       $5\times 5\times 7\times 7\times 7=5^2\times 7^3$

(v)          $2\times 2\times a\times a=2^2\times a^2$

(vi)       $a\times a\times a\times c\times c\times c\times c\times d=a^3\times c^4\times d$

Express each of the following numbers using exponential notation:

(i)             512

(ii)          343

(iii)      729

(iv)       3125

 

(i)             512
We can reduce the number 512 as given below

2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
1

So, $512=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=2^9$

(ii)          Given, 343
We can reduce the number 343 as given below

7	343
7	49
7	7
1

        So, $343=7\times 7\times 7=7^3$

(iii)      Given, 729

We can reduce the number 729 as given below

3	729
3	243
3	81
3	27
3	9
3	3
1

 

Hence, $729=3\times 3\times 3\times 3\times 3\times 3=3^6$

 

(iv)       3125

We can reduce the number 3125 as given below

5	3125
5	625
5	125
5	25
5	5
1

Hence, $3125=5\times 5\times 5\times 5\times 5=5^5$

Identify the greater number, wherever possible, in each of the following?

(i)             $4^3\text{ or }{3}^4$

(ii)          $5^3\text{ or }{3}^5$ 

(iii)      $2^8\text{ or }{8}^2$

(iv)       ${100}^2\text{ or }{2}^{100}$

(v)          $2^{10}\text{ or }{10}^2$

(i)             $4^3=4\times 4\times 4=64$

$3^4=3\times 3\times 3\times 3=81$

Since $64<81$

Thus, $3^4$ is greater than $4^3$.

(ii)          $5^3=5\times 5\times 5=125$

$3^5=3\times 3\times 3\times 3\times 3=243$

Since, $125<243$

Thus, $3^5$ is greater than $5^3$.

(iii)      $2^8=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=256$

$8^2=8\times 8=64$

Since, $256>64$

Thus, $2^8$ is greater than $8^2$.

(iv)       ${100}^2=100\times 100=10,000$

$2^{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1,024$

$2^{100}={\left(2^{10}\right)}^{10}=1,024\times 1,024\dots \mathrm{..10}\text{ times}$

Since, $10,000<1024\times \dots \dots \times 10\text{times}$

Thus, $2^{100}$ is greater than ${100}^2$.

(v)          $2^{10}=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2=1,024$

${10}^2=10\times 10=100$

Since, $1,024>100$

Thus, $2^{10}>{10}^2$

Express each of the following as product of powers of their prime factors:

(i)             648

(ii)          405

(iii)      540

(iv)       3,600

(i)             $648$

Reducing 648 in prime factors

2	648
2	324
2	162
3	81
3	27
3	9
3	3
1

$648=2^3\times 3^4$

(ii)          $405$

Reducing 405 into its prime factors.

5	405
3	81
3	27
3	9
3	3
1

$405=5\times 3^4$

(iii)      $540$

2	540
2	270
3	135
3	45
3	15
5	5
1

 

Hence, $540=2^2\times 3^3\times 5$

(iv)       $3,600$

2	3600
2	1800
2	900
2	450
3	225
3	75
5	25
5	5
1

 

$3,600=2^4\times 3^2\times 5^2$

Simplify:

(i)             $2\times {10}^3$

(ii)          $7^2\times 2^2$ 

(iii)      $2^3\times 5$

(iv)       $3\times 4^4$

(v)          $0\times {10}^2$

(vi)       $5^2\times 3^3$

(vii)    $2^4\times 3^2$

(viii)$3^2\times {10}^4$

(i)             $2\times {10}^3=2\times 10\times 10\times 10=2,000$

(ii)          $7^2\times 2^2=7\times 7\times 2\times 2=196$ 

(iii)      $2^3\times 5=2\times 2\times 2\times 5=40$

(iv)       $3\times 4^4=3\times 4\times 4\times 4\times 4=768$

(v)          $0\times {10}^2=0\times 10\times 10=0$

(vi)       $5^2\times 3^3=5\times 5\times 3\times 3\times 3=675$

(vii)    $2^4\times 3^2=2\times 2\times 2\times 2\times 3\times 3=144$

(viii)$3^2\times {10}^4=3\times 3\times 10\times 10\times 10\times 10=90,000$

Simplify:

(i)             ${\left(-4\right)}^3$ 

(ii)          $\left(–3\right)\times {(–2)}^3$

(iii)      ${\left(–3\right)}^2\times {(–5)}^2$

(iv)       ${\left(–2\right)}^3\times {(–10)}^3$

(i)             ${\left(-4\right)}^3=(-4)\times (-4)\times (-4)=-64$ 

(ii)          $\left(–3\right)\times {(–2)}^3=(-3)\times (-2)\times (-2)\times (-2)=24$

(iii)      ${\left(–3\right)}^2\times {(–5)}^2=(-3)\times (-3)\times (-5)\times (-5)=225$

(iv)       $\begin{array}{l}{\left(–2\right)}^3\times {(}^{–}\\ =(-2)\times (-2)\times (-2)\times (-10)\times (-10)\times (-10)\\ =8,000\end{array}$

Compare the following numbers:

(i)             $2.7\times {10}^{12};1.5\times {10}^8$

(ii)          $4\times {10}^{14};3\times {10}^{17}$

(i)             $2.7\times {10}^{12}\text{ and }1.5\times {10}^8$ 
On comparing the exponents of base $10$,

${10}^{12}>{10}^8$

Therefore, $2.7\times {10}^{12}\text{ >}1.5\times {10}^8$

(ii)          $4\times {10}^{14}\text{ and }3\times {10}^{17}$
On comparing the exponents of base $10$,
$4\times {10}^{14}<3\times {10}^{17}$

 

 

 

Using laws of exponents, simplify and write the answer in exponential form:

(i)             $3^2\times 3^4\times 3^8$

(ii)          $6^{15}÷6^{10}$

(iii)      $a^3\times a^2$

(iv)       $7^x\times 7^2$

(v)          ${(5^2)}^3÷5^3$

(vi)       $2^5\times 5^5$

(vii)    $a^4\times b^4$

(viii) ${\left({\text{3}}^4\right)}^3$

(ix)       $(2^{20}÷2^{15})\times 2^3$

(x)          $8^t÷8^2$

(i)             $3^2\times 3^4\times 3^8=3^{(2+4+8)}=3^{14}$       $[\because a^m\times a^n=a^{m+n}]$

(ii)          $6^{15}÷6^{10}=6^{15-10}=6^5$               $[\because a^m÷a^n=a^{m-n}]$

(iii)      $a^3\times a^2=a^{3+2}=a^5$                    $[\because a^m\times a^n=a^{m+n}]$

(iv)       $7^x\times 7^2=7^{x+2}$                          $[\because a^m\times a^n=a^{m+n}]$

(v)          $\begin{array}{l}{(}^{5^2}÷5^3=5^{2\times 3}÷5^3\\ =5^6÷5^3\\ =5^{6-3}\\ =5^3\end{array}$                $\begin{array}{l}[\because {\left(a^m\right)}^n=a^{m\times n}]\\ [\because a^m÷a^n=a^{m-n}]\end{array}$

(vi)       $2^5\times 5^5={(2\times 5)}^5={10}^5$             $[\because a^m\times b^m={(a\times b)}^m]$

(vii)    $a^4\times b^4={(a\times b)}^4$                      $[\because a^m\times b^m={(a\times b)}^m]$

(viii) ${\left({\text{3}}^4\right)}^3=3^{4\times 3}=3^{12}$                    $[\because {\left(a^m\right)}^n=a^{m\times n}]$

(ix)       $\begin{array}{l}(2^{20}÷2^{15})\times 2^3\\ =(2^{20-15})\times 2^3\\ =2^5\times 2^3\\ =2^{5+3}\\ =2^8\end{array}$                 $\begin{array}{l}\\ [\because a^m÷a^n=a^{m-n}]\\ \\ [\because a^m\times a^n=a^{m+n}]\end{array}$

(x)          $8^t÷8^2=8^{t-2}$                            $[\because a^m÷a^n=a^{m-n}]$

Simplify and express each of the following in exponential form:

(i)             $\frac{2^3\times 3^4\times 4}{3\times 32}$ 

(ii)          $\left({\left(5^2\right)}^3\times 5^4\right)÷5^7$

(iii)      ${25}^4÷5^3$

(iv)       $\frac{3\times 7^2\times {11}^8}{21\times {11}^3}$

(v)          $\frac{3^7}{3^4\times 3^3}$

(vi)       $2^0+3^0+4^0$

(vii)    $2^0\times 3^0\times 4^0$

(viii)$\left(3^0+2^0\right)\times 5^0$

(ix)       $\frac{2^8\times a^5}{4^3\times a^3}$

(x)          $\left(\frac{a^5}{a^3}\right)\times a^8$

(xi)       $\frac{4^5\times a^8{b}^3}{4^5\times a^5{b}^2}$

(xii)    ${(2^3\times 2)}^2$

(i)             $\begin{array}{l}\frac{2^3\times 3^4\times 4}{3\times 32}\\ =\frac{2^3\times 3^4\times 2^2}{3\times 2^5}\\ =\frac{2^{3+2}\times 3^4}{3\times 2^5}[\because a^m\times a^n=a^{m+n}]\end{array}$     
$\begin{array}{l}=\frac{2^5\times 3^4}{3\times 2^5}\\ =2^{5-5}\times 3^{4-1}[\because a^m÷a^n=a^{m-n}]\\ =2^0\times 3^3\\ =1\times 3^3\\ =3^3\end{array}$

(ii)          $\begin{array}{l}{25}^4÷5^3\\ ={(}^{5^2}÷5^3\\ =5^8÷5^3\end{array}$                                     $\begin{array}{l}[\because {\left(a^m\right)}^n=a^{m\times n}]\\ [\because a^m÷a^n=a^{m-n}]\end{array}$
$\begin{array}{l}=5^{8-3}\\ =5^5\end{array}$

(iii)      $\begin{array}{l}\frac{3\times 7^2\times {11}^8}{21\times {11}^3}\\ =\frac{3\times 7^2\times {11}^8}{3\times 7\times {11}^3}\end{array}$                                 
$\begin{array}{l}=3^{1-1}\times 7^{2-1}\times {11}^{8-3}[\because a^m÷a^n=a^{m-n}]\\ =3^0\times 7^1\times {11}^5\\ =7\times {11}^5\end{array}$

(iv)       $\begin{array}{l}\frac{3^7}{3^4\times 3^3}\\ =\frac{3^7}{3^{4+3}}[\because a^m\times a^n=a^{m+n}]\end{array}$    
$\begin{array}{l}=\frac{3^7}{3^7}\\ =3^{7-7}[\because a^m÷a^n=a^{m-n}]\\ =3^0\\ =1\end{array}$

(v)          $\begin{array}{l}{2}^0+3^0+4^0\\ =1+1+1\\ =3\end{array}$                                     $[\because a^0=1]$

(vi)       $\begin{array}{l}{2}^0\times 3^0\times 4^0\\ =1\times 1\times 1\\ =1\end{array}$                                      $[\because a^0=1]$

(vii)    $\begin{array}{l}\left(3^0+2^0\right)\times 5^0\\ =(1+1)\times 1\\ =2\times 1\\ =2\end{array}$                                   $[\because a^0=1]$

(viii)$\begin{array}{l}\frac{2^8\times a^5}{4^3\times a^3}\\ =\frac{2^8\times a^5}{{\left(2^2\right)}^3\times a^3}\end{array}$                                   
$=\frac{2^8\times a^5}{2^6\times a^3}[\because {\left(a^m\right)}^n=a^{m\times n}]$
$\begin{array}{l}=2^{8-6}{a}^{5-3}[\because a^m÷a^n=a^{m-n}]\\ =2^2\times a^2\\ ={(}^2[\because a^m\times b^m={(}^a]\end{array}$

(ix)       $\begin{array}{l}\left(\frac{a^5}{a^3}\right)\times a^8\\ =(a^{5-3})\times a^8[\because a^m÷a^n=a^{m-n}]\\ =a^2\times a^8\\ =a^{2+8}=a^{10}[\because a^m\times a^n=a^{m+n}]\end{array}$       

(x)          $\begin{array}{l}\frac{4^5\times a^8{b}^3}{4^5\times a^5{b}^2}\\ =4^{5-5}\times a^{8-5}\times b^{3-2}\\ =4^0\times a^3\times b\\ =1\times a^3\times b\\ =a^{3}b\end{array}$                         . $\begin{array}{l}\\ [\because a^m÷a^n=a^{m-n}]\\ [\because a^0=1]\end{array}$.

(xi)       ${(2^3\times 2)}^2$                                          

        $\begin{array}{l}={\left(2^{3+1}\right)}^2[\because a^m\times a^n=a^{m+n}]\\ ={\left(2^4\right)}^2\\ =2^{4\times 2}\\ =2^8\end{array}$

Say true or false and justify your answer:

(i)             $10\times {10}^{11}={100}^{11}$

(ii)          $2^3>5^2$

(iii)      $2^3\times 3^2=6^5$

(iv)       $3^0={\left(1000\right)}^0$

(i)             $10\times {10}^{11}={100}^{11}$

$\text{L}\text{.H}\text{.S}\text{. }{10}^{1+11}={10}^{12}$ 
and

$\text{R}\text{.H}\text{.S}\text{. }{100}^{11}\text{=}{\left({10}^2\right)}^{11}={10}^{22}$

Since, $\text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}$

Therefore, it is false.

(ii)          $2^3>5^2$

$\text{L}\text{.H}\text{.S}\text{. }{2}^3=8$ 
and $\text{R}\text{.H}\text{.S}\text{. }{5}^2=25$

Since, L.H.S. is not greater than R.H.S.

Therefore, it is false.

(iii)      $2^3\times 3^2=6^5$

$\text{L}\text{.H}\text{.S}\text{. }{2}^3\times 3^2=8\times 9=72$ 
and $\text{R}\text{.H}\text{.S}\text{. }{6}^5=7,776$

Since, $\text{L}\text{.H}\text{.S}\text{.}\ne \text{R}\text{.H}\text{.S}\text{.}$

Therefore, it is false.

(iv)       $3^0={\left(1000\right)}^0$

$\text{L}\text{.H}\text{.S}\text{. }{3}^0=1$ 
and $\text{R}\text{.H}\text{.S}\text{. }{(1000)}^0=1$

Since, $\text{L}\text{.H}\text{.S}\text{.}=\text{R}\text{.H}\text{.S}\text{.}$

Therefore, it is true.

Express each of the following as a product of prime factors only in exponential form:

(i)             $108\times 192$

(ii)          $270$

(iii)      $729\times 64$

(iv)       $768$

 

 

 

 

 

 

 

 

 

 

 

 

(i)             $108\times 192$

$\begin{array}{l}=(2^2\times 3^3)\times (2^6\times 3)\\ =2^{2+6}\times 3^{3+1}\\ =2^8\times 3^4\end{array}$

 

 

2	108
2	54
3	27
3	9
3	3
1

2	192
2	96
2	48
2	24
2	12
2	6
3	3
1

 

 

 

 

 

 

 

 

 

 

 

(ii)          $270$

2	270
3	135
3	45
3	15
5	5
1

Hence, $270=2\times 3^3\times 5$

(iii)      $729\times 64$

 

 

 

 

 

2	64
2	32
2	16
2	8
2	4
2	2
1

3	729
3	243
3	81
3	27
3	9
3	3
1

 

 

 

          

 

 

 

 

 

Hence, $729\times 64=3^6\times 2^6$

(iv)       $768$ 

2	768
2	384
2	192
2	96
2	48
2	24
2	12
2	6
3	3
1

 

Hence, $768=2^8\times 3$

Simplify:

(i)             $\frac{{\left(2^5\right)}^2\times 7^3}{8^3\times 7}$ 

(ii)          $\frac{25\times 5^2\times t^8}{{10}^3\times t^4}$

(iii)      $\frac{3^5\times {10}^5\times 25}{5^7\times 6^5}$

(i)             $\begin{array}{l}\frac{{\left(2^5\right)}^2\times 7^3}{8^3\times 7}\\ =\frac{2^{5\times 2}\times 7^3}{{(2^3)}^3\times 7}\end{array}$ 

$\begin{array}{l}=\frac{2^{10}\times 7^3}{2^9\times 7}\\ =2^{10-9}\times 7^{3-1}\end{array}$

$\begin{array}{l}=2\times 7^2\\ =2\times 49\\ =98\end{array}$

(ii)          $\begin{array}{l}\frac{25\times 5^2\times t^8}{{10}^3\times t^4}\\ =\frac{5^2\times 5^2\times t^8}{{(5\times 2)}^3\times t^4}\end{array}$
$\begin{array}{l}=\frac{5^{2+2}\times t^{8-4}}{2^3\times 5^3}\\ =\frac{5^4\times t^4}{2^3\times 5^3}\\ =\frac{5^{4-3}\times t^4}{2^3}\\ =\frac{5t^4}{8}\end{array}$

(iii)      $\begin{array}{l}\frac{3^5\times {10}^5\times 25}{5^7\times 6^5}\\ =\frac{3^5\times {(2\times 5)}^5\times 5^2}{5^7\times {(2\times 3)}^5}\end{array}$

$=\frac{3^5\times 2^5\times 5^5\times 5^2}{5^7\times 2^5\times 3^5}$

$=\frac{3^5\times 2^5\times 5^{5+2}}{5^7\times 2^5\times 3^5}$

$\begin{array}{l}=\frac{3^5\times 2^5\times 5^7}{5^7\times 2^5\times 3^5}\\ =2^{5-5}\times 3^{5-5}\times 5^{7-7}\\ =2^0\times 3^0\times 5^0\\ =1\times 1\times 1\\ =1\end{array}$

Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

(i)             $2,79,404$

$=2,00,000+70,000+9,000+400+0+4$

$=2\times 100000+7\times 10000+9\times 1000+4\times 100+0\times 10+4\times 1$

$=2\times {10}^5+7\times {10}^4+9\times {10}^3+4\times {10}^2+0\times {10}^1+4\times {10}^0$

(ii)          $30,06,194$

$=30,00,000+0+0+6,000+100+90+4$

$=3\times 1000000+0\times 100000+0\times 10000+6\times 1000+1\times 100+9\times 10+4\times 1$

$=3\times {10}^6+0\times {10}^5+0\times {10}^4+6\times {10}^3+1\times {10}^2+9\times 10+4\times {10}^0$

(iii)      $28,06,196$

$=20,00,000+8,00,000+0+6,000+100+90+6$

$=2\times 1000000+8\times 100000+0\times 10000+6\times 1000+1\times 100+9\times 10+6\times 1$

$=2\times {10}^6+8\times {10}^5+0\times {10}^4+6\times {10}^3+1\times {10}^2+9\times 10+6\times {10}^0$

(iv)       $1,20,719$

$=1,00,000+20,000+0+700+10+9$

$=1\times 100000+2\times 10000+0\times 1000+7\times 100+1\times 10+9\times 1$

$=1\times {10}^5+2\times {10}^4+0\times {10}^3+7\times {10}^2+1\times {10}^1+9\times {10}^0$

(v)          $20,068$

$=20,000+0+0+60+8$

$=2\times 10000+0\times 1000+0\times 100+6\times 10+8\times 1$

$=2\times {10}^4+0\times {10}^3+0\times {10}^2+6\times {10}^1+8\times {10}^0$

Find the number from each of the following expanded forms:

a.    $8\times {10}^4+6\times {10}^3+0\times {10}^2+4\times {10}^1+5\times {10}^0$

b.   $4\times {10}^5+5\times {10}^3+3\times {10}^2+2\times {10}^0$

c.    $3\times {10}^4+7\times {10}^2+5\times {10}^0$

d.   $9\times {10}^5+2\times {10}^2+3\times {10}^1$

a.    $8\times {10}^4+6\times {10}^3+0\times {10}^2+4\times {10}^1+5\times {10}^0$

$=8\times 10000+6\times 1000+0\times 100+4\times 10+5\times 1$

$=80000+6000+0+40+5$

$=86,045$

b.   $4\times {10}^5+5\times {10}^3+3\times {10}^2+2\times {10}^0$

$=4\times 100000+0\times 10000+5\times 1000+3\times 100+0\times 10+2\times 1$

$=400000+0+5000+300+0+2$

$=4,05,302$

c.    $3\times {10}^4+7\times {10}^2+5\times {10}^0$

$=3\times 10000+0\times 1000+7\times 100+0\times 10+5\times 1$

$=30000+0+700+0+5$

$=30,705$

d.   $9\times {10}^5+2\times {10}^2+3\times {10}^1$

$=9\times 100000+0\times 10000+0\times 1000+2\times 100+3\times 10+0\times 1$

$=900000+0+0+200+30+0$

$=9,00,230$

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

(i)             $5,00,00,000=5\times 1,00,00,000=5\times {10}^7$

(ii)          $70,00,000=7\times 10,00,000=7\times {10}^6$

(iii)      $3,18,65,00,000=31865\times 100000$

$=3.1865\times 10000\times 100000=3.1865\times {10}^9$

(iv)       $3,90,878=3.90878\times 100000=3.90878\times {10}^5$

(v)          $39087.8=3.90878\times 10000=3.90878\times {10}^4$

(vi)       $3908.78=3.90878\times 1000=3.90878\times {10}^3$

Express the number appearing in the following statements in standard form.

a.    The distance between Earth and Moon is $384,000,000\text{ m}$.

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 on 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.

a.    The distance between Earth and Moon $=384,000,000\text{ m}$

$=384\times 1000000\text{ m}$

$=3.84\times 100\times 1000000$

$=3.84\times {10}^8\text{ m}$

b.   Speed of light in vacuum
$=300,000,000\text{ m/s}$
$=3\times 100000000\text{ m/s}$
$=3\times {10}^8\text{ m/s}$

c.    Diameter of the Earth
$=1,27,56,000\text{ m}$
$=12756\times 1000\text{ m}$
$=1.2756\times 10000\times 1000\text{ m}$
$=1.2756\times {10}^7\text{ m}$

d.   Diameter of the Sun
$=1,400,000,000\text{ m}$
$=14\times 100,000,000\text{ m}$
$=1.4\times 10\times 100,000,000\text{ m}$
$=1.4\times {10}^9\text{ m}$ 

e.    Average of Stars
$=100,000,000,000$ 
$=1\times 100,000,000,000$ 
$=1\times {10}^{11}$ 

f.     Years of Universe
$=12,000,000,000\text{ years}$
$=12\times 1000,000,000\text{ years}$
$=1.2\times 10\times 1000,000,000\text{ years}$
$=1.2\times {10}^{10}\text{ years}$

g.   Distance of the Sun from the centre of the Milky Way Galaxy
$=300,000,000,000,000,000,000\text{ m}$
$=3\times 100,000,000,000,000,000,000\text{ m}$
$=3\times {10}^{20}\text{ m}$

h.   Number of molecules in a drop of water weighing $1.8$ gm
$=60,230,000,000,000,000,000,000$
$=6023\times 10,000,000,000,000,000,000$
$=6.023\times 1000\times 10,000,000,000,000,000,000$
$=6.023\times {10}^{22}$

i.      The Earth has Sea water
$=1,353,000,000{\text{ km}}^3$ 
$=1,353\times 1000000{\text{ km}}^3$ 
 

$=1.353\times 1000\times 1000,000{\text{ km}}^3$ 
$=1.353\times {10}^9{\text{ km}}^3$

j.      The population of India
$=1,027,000,000$
$=1027\times 1000000$ 
$=1.027\times 1000\times 1000000$ 
$=1.027\times {10}^9$