Unit: 13: Exponents and Powers

In questions 1 to 22, there are four options, out of which one is correct. Write the correct one.

${\left[{\left(-3\right)}^2\right]}^3$ is equal to

a.    ${\left(-3\right)}^8$

b.   ${\left(-3\right)}^6$

c.    ${\left(-3\right)}^5$

d.   ${\left(-3\right)}^{23}$

(b)

We have studied that, if $‘a’$ is a rational number, $m$ and $n$  are natural numbers, then ${(a^m)}^n =a^{m\times n}$ $a=(-3),m=2,n=3$

$\text{So, }{[{(-3)}^2]}^3 ={\left(-3\right)}^{2\times 3} ={\left(-3\right)}^6.$

For a non-zero rational number $x,x^8÷x^2$ is equal to

a.    $x^4$

b.   $x^6$

c.    $x^{10}$

d.   $x^{16}$

(b)

We have studied that, if $‘a’$ is a rational number, $m\text{ and }n$ are natural numbers such that $m>n,$ then $a^m÷a^n=a^{(m-n)}$

Here, we have $a=x,m=8,n=2$

So, $\begin{array}{c}{x}^8÷x^2=\frac{x^8}{x^2}\\ =x^{8-2}\\ =x^6\end{array}$ 

$x$ is a non-zero rational number. Product of the square of $x$ with the cube of $x$ is equal to the

a.    second power of $x$

b.   third power of $x$

c.    fifth power of $x$

d.   sixth power of $x$

(c)

Square of $x=x^2$

Cube of $x=x^3$

Product of square with the cube of $x$  $\begin{array}{l}=x^2\times x^3\\ =x^{2+3}[\therefore a^m\times a^n=a^{m+n}]\\ =x^5\end{array}$

i.e. fifth power of $x$.

For any two non-zero rational numbers $x$ and $y$, $x^5÷y^5$ is equal to

a.    ${\left(x÷y\right)}^1$

b.   ${\left(x÷y\right)}^0$

c.    ${\left(x÷y\right)}^5$

d.   ${\left(x÷y\right)}^{10}$

(c)

It is given that, $x^5÷y^5=\frac{x^5}{y^5}$

We have studied that, $\frac{p^n}{q^n}={\left(\frac{p}{q}\right)}^n$

Here, $p=x,q=y,n=5$

Thus, $\begin{array}{c}\frac{x^5}{y^5}={\left(\frac{x}{y}\right)}^5\\ ={(}^x\end{array}$

$a^m\times a^n$ is equal to

a.    ${(a^2)}^{mn}$

b.   $a^{m-n}$

c.    $a^{m+n}$

d.   $a^{mn}$

(c)

As we have studied that, if $‘a’$ is a rational number, $m$ and $n$ are natural numbers, then $a^m\times a^n=a^{m+n}$ 

$\left(1^0+2^0+3^0\right)$ is equal to

a.    $0$

b.   $1$

c.    $3$

d.   $6$

(c)

As we have studied that, $a^0=1$

$\therefore$ $\begin{array}{l}{1}^0+2^0+3^0\\ =1+1+1\\ =3\end{array}$

Value of $\frac{{10}^{22}+{10}^{20}}{{10}^{20}}$ is

a.    $10$

b.   ${10}^{42}$

c.    $101$

d.   ${10}^{22}$

(c)

 The given expression can be written as

 $\begin{array}{l}\frac{{10}^{22}}{{10}^{20}}+\frac{{10}^{20}}{{10}^{20}}\\ ={10}^{22-20}+{10}^{20-20}\end{array}$

$\begin{array}{l}={10}^2+1\\ =10\times 10+1\\ =100+1\\ =101\end{array}$

The standard form of the number $12345$ is

a.    $12345\times {10}^1$

b.   $123.45\times {10}^2$

c.    $12.345\times {10}^3$

d.   $1.2345\times {10}^4$

(d)

A number in its standard form is written as ‘a $\times {10}^k,$ where a is a terminating decimal number such that $1\le a<10$ and $k$ is any integer.

Thus, the standard form of the number $12345=1.2345\times {10}^4$

If $2^{1998}-2^{1997}-2^{1996}-2^{1995}$ $=\text{K}\text{.}{2}^{1995}$, then the value of K is

a.    $1$

b.   $2$

c.    $3$

d.   $4$

(c) It is given that

$2^{1998}–2^{1997}–2^{1996}+2^{1995}=\text{K}{.2}^{1995}$

$\Rightarrow 2^{1995}{}^{+3}-2^{1995+2}-2^{1995+1}+2^{1995}=\text{K}{.2}^{1995}$

$\Rightarrow 2^{1995}[2^3-2^2-2^1+1]=\text{K}{.2}^{1995}$

$\Rightarrow 2^{1995}[8-4-2+1]=\text{K}{.2}^{1995}$

$\Rightarrow 3=\frac{\text{K}{.2}^{1995}}{2^{1995}}$

$\Rightarrow 3=\text{K or K}=3$

Thus, the value of $\text{K}$ is $3.$

Which of the following is equal to 1?

a.    $2^0+3^0+4^0$

b.   $2^0\times 3^0\times 4^0$

c.    $\left(3^0-2^0\right)\times 4^0$

d.   $\left(3^0-2^0\right)\times \left(3^0+2^0\right)$

(b)

Let us solve all the expressions one by one,

Option (a),

$2^0+3^0+4^0=1+1+1 =3\left[\because a^0=1\right]$

Option (b),

$2^0\times 3^0\times 4^0=1\times 1\times 1 =1\left[\because a^0=1\right]$

Hence, option (b) is the answer.

In standard form, the number $72105.4$ is written as $7.21054\times {10}^n$ where $n$ is equal to

a.    $2$

b.   $3$

c.    $4$

d.   $5$

(c)

We have studied that, if the given number is greater than or equal to $10,$ then the power of $10$ (i.e. $n$ ) is a positive integer and is equal to the number of places the decimal point has been shifted.

Hence, $72105.4=7.21054\times {10}^4$   

Square of $\left(\frac{-2}{3}\right)$ is

a.    $\frac{-2}{3}$

b.   $\frac{2}{3}$

c.    $\frac{-4}{9}$

d.   $\frac{4}{9}$

(d)

As per question, square of $-\frac{2}{3} \text{ is }{\left(-\frac{2}{3}\right)}^2$

Thus, $\begin{array}{l}{\left(-\frac{2}{3}\right)}^2\because {\left(-1\right)}^2=1\\ =\left(-\frac{2}{3}\right)\times \left(-\frac{2}{3}\right)\\ =\frac{4}{9}\end{array}$

Cube of $\left(\frac{-1}{4}\right)$ is

a.    $-\frac{1}{12}$

b.   $\frac{1}{16}$

c.    $-\frac{1}{64}$

d.   $\frac{1}{64}$

(c)

As per question, cube of $(-\frac{1}{4})\text{ is }{(-\frac{1}{4})}^3$

Thus, ${\left(-\frac{\text{1}}{4}\right)}^{\text{3}}=\left(-\frac{\text{1}}{4}\right)\times \left(-\frac{\text{1}}{4}\right)\times \left(-\frac{\text{1}}{4}\right)$

$\begin{array}{l}=\frac{(-\text{1})\times (-\text{1})\times (-\text{1})}{4\times 4\times 4}\because {\left(-1\right)}^3=-1\\ =-\frac{\text{1}}{64}\end{array}$

Which of the following is not equal to ${\left(\frac{-5}{4}\right)}^4$?

a.    $$$\frac{{\left(-5\right)}^4}{4^4}$

b.   $\frac{5^4}{{\left(-4\right)}^4}$

c.    $-\frac{5^4}{4^4}$

d.   $\left(-\frac{5}{4}\right)\times \left(-\frac{5}{4}\right)\times \left(-\frac{5}{4}\right)\times \left(-\frac{5}{4}\right)$

(c)

We have studied that, ${\left(\frac{p}{q}\right)}^m=\frac{p^m}{q^m}$

So, ${\left(-\frac{5}{4}\right)}^4=\frac{{(-5)}^4}{4^4}$

$\text{or }{\left(-\frac{5}{4}\right)}^4=\frac{{(5)}^4}{{(-4)}^4}$

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

Hence, option (c) is not equal to $1.$

Which of the following is not equal to $1$?

a.    $\frac{2^3\times 3^2}{4\times 18}$

b.   $\left[{\left(-2\right)}^3\times {\left(-2\right)}^4\right]÷{\left(-2\right)}^7$

c.    $\frac{3^0\times 5^3}{5\times 25}$

d.   $\frac{2^4}{{\left(7^0+3^0\right)}^3}$

(d)

Let us solve each option one by one.

Option a,

$\begin{array}{l}\frac{2^3\times 3^2}{4\times 18}\\ =\frac{2\times 2\times 2\times 3\times 3}{4\times 18}\\ =\frac{4\times 18}{4\times 18}\\ =1\end{array}$

Option b,

$\begin{array}{l}\left[{\left(-2\right)}^3\times {\left(-2\right)}^4\right]÷{\left(-2\right)}^7\\ =\frac{\left[{(-2)}^3\times {(-2)}^4\right]}{{(-2)}^7}\end{array}$

$\begin{array}{l}=\frac{{(-2)}^{3+4}}{{(-2)}^7}\\ =\frac{{(-2)}^7}{{(-2)}^7}\\ =1\end{array}$

Option c,

$\begin{array}{l}\frac{3^0\times 5^3}{5\times 25}\\ =\frac{1\times 5\times 5\times 5}{ 5\times 25}\end{array}$

$\begin{array}{l}=\frac{5\times 25}{5\times 25}\\ =1\end{array}$

Option d,

$\begin{array}{l}\frac{2^4}{{(7^0+3^0)}^3}\\ =\frac{2^4}{{(1+1)}^3}\end{array}$

$\begin{array}{l}=\frac{2^4}{2^3} \\ =2^{4-3}\end{array}$

$\begin{array}{l}=-2^1\\ =2\end{array}$

Hence, option (d) is not equal to $1$.

${\left(\frac{2}{3}\right)}^3\times {\left(\frac{5}{7}\right)}^3$ is equal to

a.    ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^9$

b.   ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^6$

c.    ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^3$

d.   ${\left(\frac{2}{3}\times \frac{5}{7}\right)}^0$

(c)

We have studied that, if $a,b\text{ and }m$ are rational numbers, then $a^m\times b^m={(ab)}^m$

Here,

$\begin{array}{l}a=\frac{2}{3},\\ b=\frac{5}{7},\\ m=3\end{array}$

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

In standard form, the number $829030000$ is written as $\text{K}\times {10}^8$ where K is equal to

a.    $82903$

b.   $829.03$

c.    $82.903$

d.   $8.2903$

(d)

We know that, a number in a standard form is written as K $\times {10}^8,$ where $\text{K}$ is a terminating decimal such that $1\le \text{K}<10.$ 

So, there is only one option, where $\text{K}=8.2903$ less then $10.$

Which of the following has the largest value?

a.    $0.0001$

b.   $\frac{1}{10000}$

c.    $\frac{1}{{10}^6}$

d.   $\frac{1}{{10}^6}÷0.1$

(a, b)

Let us solve each option one by one.

Option a,   

$\begin{array}{l}0.0001\\ =\frac{1}{10000}\end{array}$

Option b,

$\frac{1}{\text{1}0000}$

Option c,

$\begin{array}{l}\frac{1}{{10}^6}\\ =\frac{1}{10\times 10\times 10\times 10\times 10\times 10}\\ =\frac{1}{1000000}\end{array}$

Option d,

$\begin{array}{l}\frac{1}{{10}^6}÷0.1\\ =\frac{1}{{10}^6}\times \frac{1}{0.1}\\ =\frac{1}{{10}^6}\times \frac{10}{1}\\ =\frac{10}{{10}^6}\\ =\frac{1}{{10}^5}\end{array}$

$\begin{array}{l}=\frac{1}{10\times 10\times 10\times 10\times 10}\\ =\frac{1}{ 100000}\end{array}$

The fraction whose denominator is the smallest will be the largest fraction.

Hence, $a\text{ and }b$ are the largest.

In standard form $72$ crore is written as

a.    $72\times {10}^7$

b.   $72\times {10}^8$

c.    $7.2\times {10}^8$

d.   $7.2\times {10}^7$

(c)

We have studied that, a number in standard form is written as K $\times {10}^n,$ where K is the terminating decimal such that $1\le \text{K<}10$ and $n$ is any integer.

Thus, $72$ crore $\begin{array}{l}=720000000\\ =7.2\times {10}^8\end{array}$

For non-zero numbers a and b, ${\left(\frac{a}{b}\right)}^m÷{\left(\frac{a}{b}\right)}^n$, where $m>n$, is equal to

a.    ${\left(\frac{a}{b}\right)}^{mn}$

b.   ${\left(\frac{a}{b}\right)}^{m+n}$

c.    ${\left(\frac{a}{b}\right)}^{m-n}$

d.   ${\left[{\left(\frac{a}{b}\right)}^m\right]}^n$

(c)

We have studied that, $a^m ÷a^n =a^{m-n},\left(m>n\right)$

Thus, $\begin{array}{l}{\left(\frac{a}{b}\right)}^m÷{\left(\frac{a}{b}\right)}^n\\ ={\left(\frac{a}{b}\right)}^{m-n}\end{array}$ 

Which of the following is not true?

a.    $3^2>2^3$

b.   $4^3=2^6$

c.    $3^3=9$

d.   $2^5>5^2$

(c)

Let us solve each option one by one.

Option a,

$3^2 >2^3$ 

$3\times 3>2\times 2\times 2$ 

$9>8$ (true)

 Option b,

$4^3 =2^6$ 

${(2^2)}^3 =2^6$ 

$2^6 =2^6$  (true)

Option c,

$3^{3 }=9$ 

$3\times 3\times 3=9$

$27\ne 9$ (false)

Option d,

$2^5>5^2$ 

$2\times 2\times 2\times 2\times 2=5\times 5$ 

$32>25$ (true)

Hence, option (c) is not true.

Which power of $8$ is equal to $2^6$?

a.    $3$

b.   $2$

c.    $1$

d.   $4$

(b)

Let us suppose that the power of $8$ be $x$.

According to the question, we have

$8^x=2^6$ 

${(2^3)}^x=2^6$ 

$2^{3x}=2^6$ 

Since, bases are equal, by equating their exponents, we get

$3x=6$ 

$\frac{3x}{3}=\frac{6}{3}$ 

$x=2$ 

Hence, the power of $8$ is $2,$ which is equal to $2^6$.

 

 

In questions 23 to 39, fill in the blanks to make the statements true.

 

${\left(-2\right)}^{31}\times {\left(-2\right)}^{13}={\left(-2\right)}^{?}$

Here,

 $\begin{array}{l}{\left(-2\right)}^{31}\times {\left(-2\right)}^{13}\\ \because {\text{(a)}}^{\text{m}}\times {\text{(a)}}^n={\text{a}}^{\text{m+n}}\\ ={\left(-2\right)}^{31+13}\\ ={\left(-2\right)}^{44}\\ \therefore {\left(-2\right)}^{31}\times {\left(-2\right)}^{13}\\ ={\left(-2\right)}^{44}\end{array}$ 

${\left(-3\right)}^8÷{\left(-3\right)}^5={\left(-3\right)}^{?}$

Here,

 $\begin{array}{l}{\left(-3\right)}^8÷{\left(-3\right)}^5\\ \because a^m÷a^n=a^{m-n}\\ ={\left(-3\right)}^{8-5}\\ ={\left(-3\right)}^3\\ \therefore {\left(-3\right)}^8÷{\left(-3\right)}^5\\ ={\left(-3\right)}^3\end{array}$ 

 

${\left(\frac{11}{15}\right)}^4\times {(?)}^5={\left(\frac{11}{15}\right)}^9$

Let us suppose that

${\left(\frac{11}{15}\right)}^4\times {(x)}^5={\left(\frac{11}{15}\right)}^9$ 

$\Rightarrow {(x)}^5 =\frac{{\left(\frac{11}{15}\right)}^9}{{\left(\frac{11}{15}\right)}^4}$

$\Rightarrow {(x)}^5 ={\left(\frac{11}{15}\right)}^9÷{\left(\frac{11}{15}\right)}^4$

$\Rightarrow {(x)}^5 ={\left(\frac{11}{15}\right)}^9{}^{-4}\because a^m÷a^n=a^{m-n}$

$\Rightarrow {(x)}^5 ={\left(\frac{11}{15}\right)}^5$  

Since, in the above equation, the powers are same. Thus, $x=\frac{11}{15}$ 

Hence, ${\left(\frac{11}{15}\right)}^4\times {\left(\frac{11}{15}\right)}^5$$={\left(\frac{11}{15}\right)}^9$

${\left(\frac{-1}{4}\right)}^3\times {\left(\frac{-1}{4}\right)}^{?}={\left(\frac{-1}{4}\right)}^{11}$

Let us suppose that

${\left(-\frac{1}{4}\right)}^3\times {\left(-\frac{1}{4}\right)}^x={\left(-\frac{1}{4}\right)}^{11}$

${\left(-\frac{1}{4}\right)}^x=\frac{{\left(-\frac{1}{4}\right)}^{11}}{{\left(-\frac{1}{4}\right)}^3}$

${\left(-\frac{1}{4}\right)}^x={\left(-\frac{1}{4}\right)}^{11-3}\because a^m÷a^n=a^{m-n}$

${\left(-\frac{1}{4}\right)}^x={\left(-\frac{1}{4}\right)}^8$

Since in the above equation the bases are equal. So, by equating the powers, we get, $x=8$ 

Hence, ${\left(-\frac{1}{4}\right)}^3\times {\left(-\frac{1}{4}\right)}^8={\left(-\frac{1}{4}\right)}^{11}$

${\left[{\left(\frac{7}{11}\right)}^3\right]}^4={\left(\frac{7}{11}\right)}^{?}$

Here, ${\left[{\left(\frac{7}{11}\right)}^3\right]}^4={\left(\frac{7}{11}\right)}^{3\times 4}={\left(\frac{7}{11}\right)}^{12}\because {(a^m)}^n=a^{mn}$

$\therefore {\left[{\left(\frac{7}{11}\right)}^3\right]}^4={\left(\frac{7}{11}\right)}^{12}$

${\left(\frac{6}{13}\right)}^{10}÷{\left[{\left(\frac{6}{13}\right)}^5\right]}^2={\left(\frac{6}{13}\right)}^{?}$

${\left(\frac{6}{13}\right)}^{10}÷{\left[{\left(\frac{6}{13}\right)}^5\right]}^2\because {(a^m)}^n=a^{mn}$

$={\left(\frac{6}{13}\right)}^{10}÷{\left(\frac{6}{13}\right)}^{5\times 2}$

$={\left(\frac{6}{13}\right)}^{10}÷{\left(\frac{6}{13}\right)}^{10}\because a^m÷a^n=a^{m-n}$

$={\left(\frac{6}{13}\right)}^{10-10}$

$={\left(\frac{6}{13}\right)}^0$

$\begin{array}{l}\therefore {\left(\frac{6}{13}\right)}^{10}÷{\left[{\left(\frac{6}{13}\right)}^5\right]}^2\\ ={\left(\frac{6}{13}\right)}^0\end{array}$

${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(\frac{-1}{4}\right)}^{?}$

Here,

${\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2\because {(a^m)}^n=a^{mn}$

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

$={\left(-\frac{1}{4}\right)}^{32}$

$\therefore {\left[{\left(\frac{-1}{4}\right)}^{16}\right]}^2={\left(-\frac{1}{4}\right)}^{32}$

 

${\left(\frac{13}{14}\right)}^5÷{(?)}^2={\left(\frac{13}{14}\right)}^3$

Let us suppose that

 ${\left(\frac{13}{14}\right)}^5÷{(x)}^2={\left(\frac{13}{14}\right)}^3$

${\left(\frac{13}{14}\right)}^5\times \frac{1}{{(x)}^2}={\left(\frac{13}{14}\right)}^3$

${(x)}^2=\frac{{\left(\frac{13}{14}\right)}^5}{{\left(\frac{13}{14}\right)}^3}$

${(x)}^2={\left(\frac{13}{14}\right)}^{5-3}$

${(x)}^2={\left(\frac{13}{14}\right)}^2$

Since in the above equation, the powers are same.

Thus, $x=\frac{13}{14}$

Hence, ${\left(\frac{13}{14}\right)}^5÷{\left(\frac{13}{14}\right)}^2={\left(\frac{13}{14}\right)}^3$

$a^6\times a^5\times a^0=a^{?}$

The expression can be written as $a^6\times a^5\times a^0=a^{6+5+0}=a^{11}$ 

     $\therefore a^6\times a^5\times a^0=a^{11}$ 

$1$ lakh $={10}^{?}$

We know that,

$\begin{array}{l}1\text{ lakh}\\ =100000\\ ={10}^5\end{array}$

$\therefore 1\text{ lakh}={10}^5$ 

1 million $={10}^{?}$

We know that,

$\begin{array}{l}1\text{ million}\\ =1000000\\ ={10}^6\end{array}$ 

$\therefore 1\text{ million}={10}^6$ 

$729=3^{?}$

Here, we have to find out the factors of the given expression.

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

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

$\therefore 729=3^6$

$432=2^4\times 3^{?}$

Here, we are suppose to find out the factors of given expression.

So, $432=2\times 2\times 2\times 2\times 3\times 3\times 3=2^4\times 3^3$ 

$\therefore 432=2^4\times 3^3$ 

$53700000=?\times 10^7$

The given number =   $53700000$

In standard form, it can be written as $53700000=537\times {10}^5$

Also, $537=5.37\times {10}^2\times {10}^5=5.37\times {10}^7$ 

$53700000=5.37\times {10}^7$ 

$88880000000=?\times {10}^{10}$

The given number =  $88880000000$ 

In standard form, it can be written as $88880000000=8888\times {10}^7$ 

Also, $8888=8.888\times {10}^3$ 

So, $8.888\times {10}^3\times {10}^7=8.888\times {10}^{10}$ 

$88880000000=8.888\times {10}^{10}$ 

$27500000=2.75\times {10}^{?}$

The given number =  $27500000$

This number can be expressed in standard form as

$\begin{array}{l}27500000\\ =275\times {10}^5\end{array}$ 

Also,

$\begin{array}{l}275\\ =2.75\times {10}^2\end{array}$ 

So,

$\begin{array}{l}2.75\times {10}^2\times {10}^5\\ =2.75\times {10}^7\end{array}$                    $\left[\because a^m\times a^n=a^{m+n}\right]$ 

$\begin{array}{l}27500000\\ =2.75\times {10}^{7 }\end{array}$ 

$340900000=3.409\times {10}^{?}$

The given number can be written in standard form as,

$\begin{array}{l}340900000\\ =3409\times {10}^5\end{array}$ 

Also,

$\begin{array}{l}3409\\ =3.409\times {10}^3\end{array}$ 

So,

$3.409\times {10}^3\times {10}^5$                         $\left[\because a^m\times a^n=a^{m+n}\right]$ 

$=3.409\times {10}^8$

$\begin{array}{l}340900000\\ =3.409\times {10}^8\end{array}$ 

Fill in the blanks with <, > or = sign.

a.    $3^2\_\_\_\_\_\_15$

b.   $2^3\_\_\_\_\_\_3^2$

c.    $7^4\_\_\_\_\_\_5^4$

d.   $10,000\_\_\_\_\_\_{10}^5$

e.    $6^3\_\_\_\_\_4^4$

a.    $3^2\_\_\_\_15$ 
We can write, $3^2 =3\times 3=9$ 
 So, $9<15$ 
Therefore, $3^2<15$ 

b.   $2^3\_\_\_\_3^2$ 
We can write, $2^3 =2\times 2\times 2=8$ 
And $3^2=3\times 3=9$ 
So, $8<9$ 
Therefore, $2^3<3^2$ 

c.    $7^4\_\_\_\_5^4$ 
In the above expression, as base 7 is greater than base 5 and power is same, Therefore, $7^4>5^4$ 

d.   $10000\_\_{10}^5$ 
We can write, $10000={10}^4$  
So, ${10}^4<{10}^5$ 
Therefore, $10000<{10}^5$ 

e.    $6^3\_\_\_4^4$ 
We can write, $6^3=6\times 6\times 6=216$ 
And $4^4 =4\times 4\times 4\times 4=256$
So, $216<256$ 
Therefore, $6^3<4^4$ 

 

In questions 41 to 65, state whether the given statements are True or False.

One million $={10}^7$

False

We know that, $\text{One million}=10\text{ lakhs}=1000000={10}^6$ 

Hence, ${10}^6 \ne {10}^7$ 

One hour $={60}^2$ seconds

True

We know that, $1\text{ h}=60\text{ min}=60\times 60\text{ s}={60}^2\text{ s}$ 

$1^0\times 0^1 =1$

False

We know that, $1^0\times 0^1 =1\times 0=0\ne 1$

${\left(-3\right)}^4 =-12$

False

We know that, ${\left(-3\right)}^4 =\left(-3\right)\times \left(-3\right)\times \left(-3\right)\times \left(-3\right)=81\ne -12$ 

$3^4>4^3$

True

We know that, $\because 3^4=3\times 3\times 3\times 3=81$ and

$4^3 =4\times 4\times 4=64$

$81>64$

Therefore, ${\text{3}}^{\text{4} }>{\text{4}}^{\text{3}}$

${\left(-\frac{3}{5}\right)}^{100}=\frac{-3^{100}}{-5^{100}}$

True

Considering left hand side of the equation, we have

${\left(-\frac{3}{5}\right)}^{100} ={\left(\frac{-1\times 3}{5}\right)}^{100}$

$=\frac{{\left(-1\right)}^{100}\times 3^{100}}{5^{100}}$

$=\frac{1\times 3^{100}}{5^{100}}$

$=\frac{3^{100}}{5^{100}}$

Now, taking RHS, we have $\frac{-3^{100}}{-5^{100}} =\frac{3^{100}}{5^{100}}$

LHS $=$ RHS

Hence, ${\left(-\frac{3}{5}\right)}^{100}=\frac{-3^{100}}{-5^{100}}$

${\left(10+10\right)}^{10}={10}^{10}+{10}^{10}$

False

Let us solve LHS,

${\left(10+10\right)}^{10}={(20)}^{10}={(2\times 10)}^{10}=2^{10}\times {10}^{10}$

Let us solve RHS,

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

Now,

$2^{10}\times {10}^{10}\ne 2\times {10}^{10}$

Hence, LHS is not equal to RHS

 

$x^0\times x^0=x^0÷x^0$ is true for all non-zero values of $x$.

True

As we know that

$x^0\times x^0=1\times 1=1$                  $[\because a^0=1]$ and

$x^0÷x^0=1÷1=1$                 $\left[\because a^0=1\right]$

Hence, $x^0\times x^0=x^0÷x^0$

In the standard form, a large number can be expressed as a decimal number between $0$ and $1$, multiplied by a power of $10$.

False

We know that, a number in standard form is written as $a\times {10}^k,$ where $1\le a<10\text{ and }k$ is any integer.

$4^2$ is greater than $2^4.$ 

False

$4^2=4\times 4=16$         $[\because a^m=a\times a\times a\times \dots \times a(m\text{ times})]$ and

$2^4=2\times 2\times 2\times 2=16$   

Therefore, $4^2 =2^4$

 $x^m+x^m=x^2{}^m$ where $x$ is a non-zero rational number and $m$ is a positive integer.

False

We have studied that

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

$x^m\times x^m=x^{m+m}=x^{2m}$

Also, $a^k+a^k=2a^k$

So, $x^m+x^m=2x^m\ne x^{2m}$

$x^m\times y^m={(x\times y)}^2{}^m$, where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.

False

We have studied that if $a\text{ and }b$ are rational numbers, then $x^m\times y^m={(xy)}^m$

$x^m\times y^m={(xy)}^m={(x\times y)}^m$

Hence, $x^m\times y^m\ne {(x\times y)}^{2m}$

$x^m÷y^m={(x÷y)}^m$ where $x$ and $y$ are non-zero rational numbers and $m$ is a positive integer.

True

We have studied that, if $x\text{ and }y$ are rational numbers, then

$\left(\frac{x^m}{y^m}\right)={\left(\frac{x}{y}\right)}^m=\text{or }{x}^m÷y^m={(x÷y)}^m$

$x^m\times x^n=x^m{}^{+n}$ where $x$ is a non-zero rational number and $m,n$ are positive integers.

True

We have studied that, if $a$ is a rational number and $m$ and $n$ are positive integers, then

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

$\therefore x^m\times x^n=x^{m+n}$

$4^9$ is greater than ${16}^3.$

True

$\because 16=4\times 4=4^2$              $\therefore {16}^3={\left(4^2\right)}^3=4^6$           

Now, in $4^9 \text{and }{4}^6,$$4^9>4^6$ as powers $9>6$

${\left(\frac{2}{5}\right)}^3÷{\left(\frac{5}{2}\right)}^3=1$

False

Taking LHS we get,      ${\left(\frac{2}{5}\right)}^3÷{\left(\frac{5}{2}\right)}^3={\left(\frac{2}{5}\right)}^3\times {\left(\frac{2}{5}\right)}^3$

$={\left(\frac{2}{5}\right)}^{3+3}={\left(\frac{2}{5}\right)}^6$

Hence, ${\left(\frac{2}{5}\right)}^3÷{\left(\frac{5}{2}\right)}^3\ne 1$

${\left(\frac{4}{3}\right)}^5\times {\left(\frac{5}{7}\right)}^5={\left(\frac{4}{3}+\frac{5}{7}\right)}^5$

False

Taking LHS we get,      ${\left(\frac{4}{3}\right)}^5\times {\left(\frac{5}{7}\right)}^5={\left(\frac{4}{3}\times \frac{5}{7}\right)}^5={\left(\frac{20}{21}\right)}^5$

And      ${\left[\left(\frac{4}{3}\right)+\left(\frac{5}{7}\right)\right]}^5={\left(\frac{4}{3}+\frac{5}{7}\right)}^5={\left(\frac{43}{21}\right)}^5$

As the base is not same for LHS value,

So, LHS $\ne$ RHS

Hence, ${\left(\frac{4}{3}\right)}^5\times {\left(\frac{5}{7}\right)}^5\ne {\left(\frac{4}{3}+\frac{5}{7}\right)}^5$

${\left(\frac{5}{8}\right)}^9÷{\left(\frac{5}{8}\right)}^4={\left(\frac{5}{8}\right)}^4$

False

Taking LHS we get, ${\left(\frac{5}{8}\right)}^9÷{\left(\frac{5}{8}\right)}^4={\left(\frac{5}{8}\right)}^{9-4}={\left(\frac{5}{8}\right)}^5\ne {\left(\frac{5}{8}\right)}^4$

Hence, ${\left(\frac{5}{8}\right)}^9÷{\left(\frac{5}{8}\right)}^4\ne {\left(\frac{5}{8}\right)}^4$

${\left(\frac{7}{3}\right)}^2\times {\left(\frac{7}{3}\right)}^5={\left(\frac{7}{3}\right)}^{10}$

False

Taking LHS we get, ${\left(\frac{7}{3}\right)}^2\times {\left(\frac{7}{3}\right)}^5={\left(\frac{7}{3}\right)}^{2+5}$

$={\left(\frac{7}{3}\right)}^7\ne {\left(\frac{7}{3}\right)}^{10}$

Hence, ${\left(\frac{7}{3}\right)}^2\times {\left(\frac{7}{3}\right)}^5\ne {\left(\frac{7}{3}\right)}^{10}$

$5^0\times {25}^0\times {125}^0={(5^0)}^6$

True

Taking LHS we get,

$\begin{array}{l}{5}^0\times {25}^0\times {125}^0\\ =1\times 1\times 1\left(\because a^0=1\right)\\ =1\end{array}$

 

Now, taking RHS we get,

$\begin{array}{l}{\left(5^0\right)}^6={\left(1^{}\right)}^6\left(\because a^0=1\right)\\ =1\end{array}$

Hence, $5^0\times {25}^0\times {125}^0={\left(5^0\right)}^6$

$\begin{array}{l}876543=8\times {10}^5+7\times {10}^4+6\times {10}^3+5\times {10}^2\\ +4\times {10}^1+3\times {10}^0\end{array}$

True

Considering the right hand side of the given equation, we get $8\times {10}^5+7\times {10}^4+6\times {10}^3+5\times {10}^2$ $+4\times {10}^1+3\times {10}^0$

$=8\times 100000+7\times 10000+6\times 1000+5\times 100$

$+4\times 10+3\times 1[\because a^0=1]$

$=800000+70000+6000+500+40+3$

$=876543=\text{LHS}$

Hence, RHS $=$ LHS

$600060=6\times {10}^5+6\times {10}^2$

False

Considering the right hand side of the given equation, we get  $=6\times {10}^5+6\times {10}^2=6\times 100000+6\times 100$

$=600000+600=600600\ne 600060$

Hence, $\text{RHS}\ne \text{LHS}$

$4\times {10}^5+3\times {10}^4+2\times {10}^3+1\times {10}^0=432010$

False

Considering the left hand side of the given equation, we get $=\text{4}\times \text{1}{0}^{\text{5}}+\text{3}\times \text{1}{0}^{\text{4}}+\text{2}\times \text{1}{0}^{\text{3}}+\text{1}\times \text{1}{0}^0$

$=4\times 100000+3\times 10000+2\times 1000+1\times 1$   $[\because a^0=1]$

$=400000+30000+2000+1=432001\ne 432010$

Hence, $\text{LHS}\ne \text{RHS}$

$8\times {10}^6+2\times {10}^4+5\times {10}^2+9\times {10}^0=8020509$

True

Considering the left hand side of the given equation, we get $=8\times {10}^6+2\times {10}^4+5\times {10}^2+9\times {10}^0$

$=8\times 1000000+2\times 10000+5\times 100+9\times 1$   $[\because a°=1]$

$=8000000+20000+500+9=8020509=\text{RHS}$

Hence, $\text{LHS}=\text{RHS}$

$4^0+5^0+6^0={(4+5+6)}^0$

False

Considering the left hand side of the given equation, we get, $4^0+5^0+6^0=1+1+1=3$    $[\because a^0=1]$ 
and ${\left(4+5+6\right)}^0={\left(15\right)}^0=1$

Hence, $4^0+5^0+6^0\ne {(4+5+6)}^0$

Arrange in ascending order:

$2^5,3^3,2^3\times 2,{(3^3)}^2,3^5,4^0,2^3\times 3^1$

Ascending order means arranging the numbers from least to greatest.

We have, $2^5=2\times 2\times 2\times 2\times 2=32.$

$3^3=3\times 3\times 3=27.$

$2^3\times 2=2\times 2\times 2\times 2=16.$

${(3^3)}^2=3^6[\because {(a^m)}^n=a^{mn}]$

$3^6=3\times 3\times 3\times 3\times 3\times 3=729.$

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

$4°=1 [\because a^0=1]$

and $2^3\times 3=2\times 2\times 2\times 3=24$

Thus, the required ascending order will be

$4^0<2^3 \times 2<2^3\times 3^1<3^3<2^5<3^5<{\left(3^3\right)}^2.$

Arrange in descending order:

$2^{2+3},{(2^2)}^3,2\times 2^2,\frac{3^5}{3^2},3^2\times 3^0,2^3\times 5^2$

Descending order means arranging the numbers from greatest to least.

We have, $2^{2+3}=2^5=2\times 2\times 2\times 2\times 2=32.$

${(2^2)}^3=2^6=2\times 2\times 2\times 2\times 2\times 2=64.$

$2\times 2^{2 }=2^{1+2}=2^3=8.$

$\frac{3^5}{3^2} =3^{5-2}=3^3=27.$

$3^2\times 3^0=3^{2+0}=3^2=9.$

$2^3\times 5^2=2\times 2\times 2\times 5\times 5$

$=8\times 25=200.$

Thus, the required descending order will be

$(2^3\times 5^2)>{(2^2)}^3>2^{2+3}>\frac{3^5}{3^2}>(3^2\times 3^0)>(2\times 2^2)$

By what number should ${\left(-4\right)}^5$ be divided so that the quotient may be equal to ${\left(-4\right)}^3$?

In order to find the number, that should divide ${\left(–4\right)}^5$ to get the quotient ${\left(–4\right)}^3,$ we will divide ${(–4)}^5$ by ${\left(–4\right)}^3.$ $={\left(–4\right)}^{5-3}={(-4)}^2$

Hence, the required number is ${(-4)}^2$

Find $m$ so that ${\left(\frac{2}{9}\right)}^3\times {\left(\frac{2}{9}\right)}^6={\left(\frac{2}{9}\right)}^{2m-1}$ 

We have,

${\left(\frac{2}{9}\right)}^3\times {\left(\frac{2}{9}\right)}^6={\left(\frac{2}{9}\right)}^{2m-1}$

$\Rightarrow {\left(\frac{2}{9}\right)}^{3+6}={\left(\frac{2}{9}\right)}^{2m-1}$

$\Rightarrow {\left(\frac{2}{9}\right)}^9={\left(\frac{2}{9}\right)}^{2m-1}\because a^m\times a^n=a^{m+n}$

When the bases are same, we can equate the powers.

Therefore, $9=2m-1$

$9+1=2m$

$10=2m$

$\frac{10}{2}=\frac{2m}{2}$

$5=m$

Hence, $m=5$

If $\frac{p}{q}={\left(\frac{3}{2}\right)}^2÷{\left(\frac{9}{4}\right)}^0$ find the value of ${\left(\frac{p}{q}\right)}^3.$

Considering the given equation, $\frac{p}{q}={\left(\frac{3}{2}\right)}^2÷{\left(\frac{9}{4}\right)}^0$

$\frac{p}{q}={\left(\frac{3}{2}\right)}^2÷\frac{1}{1}$

$\frac{p}{q}={\left(\frac{3}{2}\right)}^2$

$\frac{p}{q}=\frac{3^2}{2^2}$

$\frac{p}{q}=\frac{9}{4}$

Cubing both sides, we get

${\left(\frac{p}{q}\right)}^3={\left(\frac{9}{4}\right)}^3$

$\begin{array}{l}{\left(\frac{p}{q}\right)}^3=\frac{9\times 9\times 9}{4\times 4\times 4}\\ =\frac{729}{64}\end{array}$

Find the reciprocal of the rational number ${\left(\frac{1}{2}\right)}^2÷{\left(\frac{2}{3}\right)}^3$

Considering the given expression

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

$\begin{array}{l}=\frac{\frac{{(1)}^2}{{(2)}^2}}{\frac{{(2)}^3}{{(3)}^3}}\\ =\frac{\frac{1}{4}}{\frac{8}{27}}\end{array}$

$\begin{array}{l}=\frac{1}{4}\times \frac{27}{8}\\ =\frac{27}{4\times 8}\\ =\frac{27}{32}\end{array}$

The reciprocal is $\frac{32}{27}$

Find the value of:

a.    $7^0$

b.   $7^7÷7^7$

c.    ${(-7)}^{2\times 7-6-8}$

d.   $(2^0+3^0+4^0)(4^0-3^0-2^0)$

e.    $2\times 3\times 4÷2^0\times 3^0\times 4^0$

f.$(8^0-2^0)\times (8^0+2^0)$

a.    $7^0=1$, as we know that any number to the power of zero is one.

b.   $\begin{array}{l}{7}^{7 }÷7^7\\ =\frac{7^7}{7^7}\\ =7^{7-7}\end{array}$

$\begin{array}{l}=7^0\\ =1\end{array}$

c.    $\begin{array}{l}{\left(-7\right)}^{2\times 7-6-8}\\ ={(}^{-}\\ ={\left(-7\right)}^0\\ =1\end{array}$

d.   $\begin{array}{l}(2^0+3^0+4^0)(4^0-3^0-2^0)\\ =(1+1+1)(1-1-1)\end{array}$

$\begin{array}{l}=\left(3\right)(-1)\\ =-3\end{array}$

e.    $\begin{array}{l}2\times 3\times 4÷2^0\times 3^0\times 4^0\\ =2\times 3\times 4÷1\times 1\times 1\end{array}$

$\begin{array}{l}=\frac{2\times 3\times 4}{1\times 1\times 1}\\ =2\times 3\times 4\\ =24\end{array}$

f.     $\begin{array}{l}(8^0-2^0)\times (8^0+2^0)\\ =(1-1)\times \left(1+1\right)\\ =0\times 2=0\end{array}$

Find the value of $n$, where n is an integer and

$2^{n-5}\times 6^{2n-4}\text{=}\frac{1}{{12}^4\times 2}\text{. }$

Considering the given equation,    

$2^{n-5}\times 6^2{}^{n-4}=\frac{1}{{12}^4\times 2}$

$\frac{2^n}{2^5}\times \frac{6^{2n}}{6^4}=\frac{1}{{12}^4\times 2}$

$\frac{2^n\times 6^{2n}}{2^5\times 6^4}=\frac{1}{{(2\times 6)}^4\times 2}$

$2^n\times {(6^2)}^n=\frac{(2^5\times 6^4)}{2^4\times 6^4\times 2}$

$2^n\times {36}^n=\frac{2^5\times 6^4}{2^5\times 6^4}$

$2^n\times {36}^n=1$

${\left(2\times 36\right)}^n=1$

${72}^n={72}^0$

$n=0$

Express the following in usual form:

a.    $8.01\times {10}^7$

b.   $1.75\times {10}^{-3}$

a.    The given number $8.01\times {10}^7$ can be written as

$\begin{array}{l}=\frac{801}{100}\times 100000000\\ =80100000\end{array}$

b.   The given number $1.75\times {10}^{-3}$ can be written as

$\begin{array}{l}=\frac{175}{100}\times \frac{1}{{10}^3}\\ =\frac{175}{100000}\\ =0.00175\end{array}$

Find the value of

a.    $2^5$

b.   $\left(-3^5\right)$

c.    $-{\left(-4\right)}^4$

We have studied that, $a^n=a\times a\times a\times \dots \dots \times a \left(n\text{ times}\right)$

a.     $\begin{array}{l}{2}^5\\ =2\times 2\times 2\times 2\times 2\\ =32\end{array}$

b.   $\begin{array}{l}{(}^{-}\\ ={(}^{-}\times 3^5\\ =-1\times 3\times 3\times 3\times 3\times 3\\ =-243\end{array}$

c.    $\begin{array}{l}-{(}^{-}\\ =-[(-4)\times (-4)\times (-4)\times (-4)]\end{array}$

$\begin{array}{l}=-[{(}^{-}(4\times 4\times 4\times 4)]\\ =-\left(256\right)\\ =-256\end{array}$

Express the following in exponential form:

a.    $3\times 3\times 3\times a\times a\times a\times a$

b.   $a\times a\times b\times b\times b\times c\times c\times c\times c$

c.    $s\times s\times t\times t\times s\times s\times t$

We have studied that, $a^n=a\times a\times a\times \mathrm{......}\times a(n\text{ times})=a^n$

a.    $\begin{array}{l}3\times 3\times 3\times a\times a\times a\times a\\ =3^3·a^4\end{array}$

b.   $\begin{array}{l}a\times a\times b\times b\times b\times c\times c\times c\times c\\ =a^2·b^3·c^4\end{array}$

c.    $\begin{array}{l}s\times s\times t\times t\times s\times s\times t\\ =s\times s\times s\times s\times t\times t\times t\\ =s^4·t^3\end{array}$

How many times of $30$ must be added together to get a sum equal to ${30}^7$?

Let $n$ be the number of times that $30$ must be added together to get a sum equal to ${30}^7.$

Therefore, we can write the above statement as -  

$30+30+\dots +30={30}^7$         $\left(n\text{ times}\right)$

$30\times n={30}^7$

$\frac{30\times n}{30}=\frac{{30}^7}{30}$

$n={30}^{7-1}$

$n={30}^6$

Hence, if $30$ is added ${30}^6$ times, then we get ${30}^7.$

Express each of the following numbers using exponential notations:

a.    $1024$

b.   $1029$

c.    $\frac{144}{875}$

a.    The number 1024 using prime factorisation can be written as -
$1024=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$ $=2^{10}$

b.   The number 1029 using prime factorisation can be written as -
$\begin{array}{l}=3\times 7\times 7\times 7\\ =3\times 7^3\end{array}$

c.    The number  $\frac{144}{875}$ using prime factorisation can be written as -
 
$\begin{array}{l}=\frac{2\times 2\times 2\times 2\times 3\times 3}{5\times 5\times 5\times 7}\\ =\frac{2^4\times 3^2}{5^3\times 7^1}\end{array}$

Identify the greater number, in each of the following:

a.    $2^6\text{ or}{6}^2$

b.   $2^9\text{or }{9}^2$

c.    $7.9\times {10}^4\text{or}5.28\times {10}^5$

a.    We know that, $2^6=2\times 2\times 2\times 2\times 2\times 2=64$ and $6^2=6\times 6=36$ 
So, ${\text{2}}^{\text{6}}>{\text{6}}^{\text{6}}$

b.   We know that, $2^9=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2\times 2$

$=512$ and $9^2=9\times 9=81$

So, $2^9>9^2$

c.    We know that, $7.9\times {10}^4=7.9\times 10000=79000\text{ and}$

$5.28=5.28\times 100000=528000$

So, $5.28\times {10}^5>7.9\times {10}^4$

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

a.    $9000$

b.   $2025$

c.    $800$

The number 9000 using prime factorisation can be written as -

$9000=2^3\times 3^2\times 5^3$

The number 2025 using prime factorisation can be written as -

The number 800 using prime factorisation can be written as -

Express each of the following in single exponential form:

a.    $2^3\times 3^3$

b.   $2^4\times 4^2$

c.    $5^2\times 7^2$

d.   ${(-5)}^5\times (-5)$

e.    ${(-3)}^3\times {(-10)}^3$

f.${\left(-11\right)}^2\times {\left(-2\right)}^2$

a.    We have,
$\begin{array}{l}{2}^3\times 3^3\\ ={(}^2\end{array}$

b.   We have,
$\begin{array}{l}{2}^4\times 4^2\\ =2^4\times {(}^{2^2}\end{array}$

c.    We have,

$\begin{array}{l}{5}^2\times 7^2\\ ={(}^5.\end{array}$ 

$={35}^2$

d.   We have,

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

$\begin{array}{l}={(}^{-}\\ ={(}^{-}\times {(}^5\\ =1\times 5^6\\ =5^6\end{array}$

e.    We have,
$\begin{array}{l}{(}^{-}\times {(}^{-}\\ ={[}^{(}\end{array}$

$={(30)}^3$

f.     We have,
$\begin{array}{l}{(}^{-}\times {(}^{-}\\ ={[}^{(}\end{array}$

Express the following numbers in standard form:

a.    $76,47,000$

b.   $8,19,00,000$

c.    $5,83,00,00,00,000$

d.   $24$ billion

a.    We have, $76,47,000=7647000.00$

We have studied that, a number in standard form is written as a $\times {10}^k,$

where $a$ is the terminating decimal such that $1\le a<10\text{ and }k$ is any integer.

So, $7647000=7647\times {10}^3=7.647\times {10}^3\times {10}^3$

$=7.647\times {10}^6$

Similarly,

b.   $8,19,00,000=81900000.00=819\times {10}^5$

$=8.19\times {10}^2\times {10}^5=8.19\times {10}^7$

c.    $5,83,00,00,00,000=583000000000.00$

$=583\times {10}^9=5.83\times {10}^2\times {10}^9=5.83\times {10}^{11}$

d.   $24$ $\text{billion}$ $=24,00,00,00,000=24\times {10}^9$

$=2.4\times {10}^1\times {10}^9=2.4\times {10}^{10}$

The speed of light in vacuum is $3\times {10}^8\text{ m/s}$. Sunlight takes about $8$ minutes to reach the earth. Express distance of Sun from Earth in standard form.

It is given that, speed of light $=3\times {10}^8$ m/s

Since the speed is given in m/s, we need to convert the time taken into seconds.

Time taken by light to reach the Earth $=8\min =8$

$\times 60s=480s\left[\because 1\min =60s\right]$

We have studied that, Distance $=$ Speed $\times$ Time $=3\times {10}^8\times 480=1440\times {10}^8$

$=1.440\times {10}^3\times {10}^8=1.44\times {10}^{11}$  $[\because {10}^3\times {10}^8={10}^{11}]$

Hence, the distance of Sun from the Earth is $1.44\times {10}^{11}m$

Simplify and express each of the following in exponential form:

a.    $\left[{\left(\frac{3}{7}\right)}^4\times {\left(\frac{3}{7}\right)}^5\right]÷{\left(\frac{3}{7}\right)}^7$

b.   $\left[{\left(\frac{7}{11}\right)}^5÷{\left(\frac{7}{11}\right)}^2\right]\times {\left(\frac{7}{11}\right)}^2$

c.    ${(3^7÷3^5)}^4$

d.   $\left(\frac{a^6}{a^4}\right)\times a^5\times a^0$

e.    $\left[{\left(\frac{3}{5}\right)}^3\times {\left(\frac{3}{5}\right)}^8\right]÷\left[{\left(\frac{3}{5}\right)}^2\times {\left(\frac{3}{5}\right)}^4\right]$

f.$(5^{15}÷5^{10})\times 5^5$

a.    We have, $\left[{\left(\frac{3}{7}\right)}^4\times {\left(\frac{3}{7}\right)}^5\right]÷{\left(\frac{3}{7}\right)}^7$
$={\left(\frac{3}{7}\right)}^{4+5}÷{\left(\frac{3}{7}\right)}^7$
$\begin{array}{l}={\left(\frac{3}{7}\right)}^9÷{\left(\frac{3}{7}\right)}^7\\ =\frac{{\left(\frac{3}{7}\right)}^9}{{\left(\frac{3}{7}\right)}^7}\\ ={\left(\frac{3}{7}\right)}^{9-7}\end{array}$
$={\left(\frac{3}{7}\right)}^2$

b.   We have, $\left[{\left(\frac{7}{11}\right)}^5÷{\left(\frac{7}{11}\right)}^2\right]\times {\left(\frac{7}{11}\right)}^2$
$\begin{array}{l}=\left[\frac{{\left(\frac{7}{11}\right)}^5}{{\left(\frac{7}{11}\right)}^2}\right]\times {\left(\frac{7}{11}\right)}^2\\ ={\left(\frac{7}{11}\right)}^{5-2}\times {\left(\frac{7}{11}\right)}^2\end{array}$
$\begin{array}{l}={\left(\frac{7}{11}\right)}^3\times {\left(\frac{7}{11}\right)}^2\\ ={\left(\frac{7}{11}\right)}^{3+2}\end{array}$
$={\left(\frac{7}{11}\right)}^5$

c.    We have, ${(3^7÷3^5)}^4$
$={(3^{7-5})}^4$
$\begin{array}{l}={\left(3^2\right)}^4\\ =3^{2\times 4}\end{array}$
$=3^8$

d.   We have, $\left(\frac{a^6}{a^4}\right)\times a^5\times a^0$
$=\left(a^{6-4}\times a^5\times 1\right)$
$\begin{array}{l}=a^2\times a^5\\ =a^{2+5}\end{array}$
$=a^7$

e.    We have, $\left[{\left(\frac{3}{5}\right)}^3\times {\left(\frac{3}{5}\right)}^8\right]÷\left[{\left(\frac{3}{5}\right)}^2\times {\left(\frac{3}{5}\right)}^4\right]$
$={\left(\frac{3}{5}\right)}^{3+8}÷{\left(\frac{3}{5}\right)}^{2+4}$
$\begin{array}{l}={\left(\frac{3}{5}\right)}^{11}÷{\left(\frac{3}{5}\right)}^6\\ =\frac{{\left(\frac{3}{5}\right)}^{11}}{{\left(\frac{3}{5}\right)}^6}\\ ={\left(\frac{3}{5}\right)}^{11-6}\end{array}$
$={\left(\frac{3}{5}\right)}^5$

f.We have, $\left(5^{15}÷5^{10}\right)\times 5^5$
$=\left(\frac{5^{15}}{5^{10}}\right)\times 5^5$

$\begin{array}{l}=5^5\times 5^5\\ =5^{5+5}\\ =5^{10}\end{array}$

Evaluate

a.    $\frac{7^8\times a^{10}{b}^7{c}^{12}}{7^6\times a^8{b}^4{c}^{12}}$

b.   $\frac{5^4\times 7^4\times 2^7}{8\times 49\times 5^3}$

c.    $\frac{125\times 5^2\times a^7}{{10}^3\times a^4}$

d.   $\frac{3^4\times {12}^3\times 36}{2^5\times 6^3}$

e.    ${\left(\frac{6\times 10}{2^2\times 5^3}\right)}^2\times \frac{25}{27}$

f.$\frac{{15}^4\times {18}^3}{3^3\times 5^2\times {12}^2}$

g.   $\frac{6^4\times 9^2\times {25}^3}{3^2\times 4^2\times {15}^6}$

a.    Considering the expression, $\frac{7^8\times a^{10}{b}^7{c}^{12}}{7^6\times a^8{b}^4{c}^{12}}$

$=\left(\frac{7^8}{7^6}\right)\times \left(\frac{a^{10}}{a^8}\right)\times \left(\frac{b^7}{b^4}\right)\times \left(\frac{c^{12}}{c^{12}}\right)$

$=7^{8-6}\times a^{10-8}\times b^{7-4}\times c^{12-12}$

$\begin{array}{l}=7^2\times a^2\times b^3\times c^0\\ =49a^2{b}^3\end{array}$

b.      Considering the expression, $\frac{5^4\times 7^4\times 2^7}{8\times 49\times 5^3}$

$=\frac{5^4\times 7^4\times 2^7}{2^3\times 7^2\times 5^3}$

$\begin{array}{l}=\left(\frac{5^4}{5^3}\right)\times \left(\frac{2^7}{2^3}\right)\times \left(\frac{7^4}{7^2}\right)\\ =5^{4-3}\times 2^{7-3}\times 7^{4-2}\end{array}$

$\begin{array}{l}=5\times 2^4\times 7^2\\ =5\times 16\times 49\end{array}$

$=3920$

c.    Considering the expression, $\frac{125\times 5^2\times a^7}{{10}^3\times a^4}$

$=\frac{5^3\times 5^2\times a^7}{{\left(2\times 5\right)}^3\times a^4}$

$=\frac{5^{3+2}\times a^7}{2^3\times 5^3\times a^4}$

$=\frac{5^5\times a^7}{2^3\times 5^3\times a^4}$

$=\left(\frac{5^5}{5^3}\right)\times \left(\frac{a^7}{a^4}\right)\times \left(\frac{1}{2^3}\right)$

$=\left(\frac{5^{5-3}\times a^{7-4}}{2^3}\right)$

$\begin{array}{l}=\frac{5^2\times a^3}{2^3}\\ =\frac{25a^3}{8}\end{array}$

d.   Considering the expression, $\frac{3^4\times {12}^3\times 36}{2^5\times 6^3}$

$=\frac{3^4\times \left(2^6\times 3^3\right)\times \left(2^2\times 3^2\right)}{2^5\times {\left(2\times 3\right)}^3}$

$=\frac{3^4\times 2^6\times 3^3\times 2^2\times 3^2}{2^5\times 2^3\times 3^3}$

$\frac{\left(3^4\times 3^3\times 3^2\right)\times \left(2^6\times 2^2\right)}{2^5\times 2^3\times 3^3}$

$=\frac{3^{4+3+2}\times 2^{6+2}}{2^{5+3}\times 3^3}$

$=\frac{3^9\times 2^8}{3^3\times 2^8}=3^{9-3}\times 2^{8-8}$

$\begin{array}{l}=3^6\times 2^0=3^6\times 1\\ =729\end{array}$

e.    Considering the expression, ${\left(\frac{6\times 10}{2^2\times 5^3}\right)}^2\times \frac{25}{27}$

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

$\begin{array}{l}=\frac{3^2}{5^4}\times \frac{5^2}{3^3}\\ =\frac{1}{5^2\times 3}\\ =\frac{1}{25\times 3}\\ =\frac{1}{75}\end{array}$

f.     Considering the expression,, $\frac{{15}^4\times {18}^3}{3^3\times 5^2\times {12}^2}$

$=\frac{{\left(3\times 5\right)}^4\times {\left(2\times 3^2\right)}^3}{3^3\times 5^2\times {\left(2^2\times 3\right)}^2}$

$=\frac{3^{4+6-3-2}\times 5^4{}^{-2}}{2^{4-3}}$

$\begin{array}{l}=\frac{3^5\times 5^2}{2}\\ =\frac{\left(243\times 25\right)}{2}\end{array}$

$=\frac{6075}{2}$

g.   Considering the expression,, $=\frac{6^4\times 9^2\times {25}^3}{3^2\times 4^2\times {15}^6}$

$=\frac{{\left(2\times 3\right)}^4\times {\left(3^2\right)}^2\times {\left(5^2\right)}^3}{3^2\times 2^4\times 3^6\times 5^6}$

$=2^{4-4}\times 3^{4+4-2-6}\times 5^{6-6}$

$=2^0\times 3^0\times 5^0$

$\begin{array}{l}=1\times 1\times 1\\ =1\end{array}$

Express the given information in Scientific notation (standard form) and then arrange them in ascending order of their size.

Sl.No.	Deserts of the World	Area (Sq. Kilometres)
(1)	Kalahari, South Africa	$932,400$
(2)	Thar, India	$199,430$
(3)	Gibson, Australia	$155,400$
(4)	Great Victoria, Australia	$647,500$
(5)	Sahara, North Africa	$8,598,800$

1.   Area of Kalahari, South Africa   

$=932,400=932400.00$ 

[Since we know that the standard form is written as $a\times {10}^k$ ]

     $\begin{array}{l}=9324\times {10}^2\\ =9.324\times {10}^3\times {10}^2\end{array}$ 

     $=9.324\times {10}^5$

2.   Area of Thar, India $=199,430=199430.00$

$\begin{array}{l}=19943\times {10}^1\\ =1.9943\times {10}^4\times {10}^1\\ =1.9943\times {10}^5\end{array}$

3.   Area of Gibson, Australia $=155,400$

$=155400.00$

$\begin{array}{l}=1554\times {10}^2\\ =1.554\times {10}^3\times {10}^2\\ =1.554\times {10}^5\end{array}$

4.   Area of Great Victoria, Australia $$$=647,500$

$\begin{array}{l}=6475\times {10}^2\\ =6.475\times {10}^3\times {10}^5\\ =6.475\times {10}^5\end{array}$

5.   Area of Sahara, North-Africa $=8,598,800$

$\begin{array}{l}=85988\times {10}^2\\ =8.5988\times {10}^4\times {10}^2\end{array}$

$=8.5988\times {10}^6$

We have studied that, to compare two numbers written in scientific notation: The number with the larger power of $10$ is greater than the number with the smaller power of $10$. If the powers of ten are the same, then the number with larger factor is the larger number. Hence, the ascending order of the sizes of the deserts will be Gibson, Australia $<$ Thar, India $<$ Great Victoria, Australia $<$ Kalahari, South-Africa $<$ Sahara, North-Africa.

Express the given information in Scientific notation and then arrange them in descending order of their size.

Sl.No.	Name of the Planet	Mass (in kg)
(1)	Mercury	$330000000000000000000000$
(2)	Venus	$4870000000000000000000000$
(3)	Earth	$5980000000000000000000000$
(4)	Mars	$642000000000000000000000$
(5)	Jupiter	$\begin{array}{l}190000000000000000000000\\ 0000\end{array}$
(6)	Saturn	$\begin{array}{l}56900000000000000000000\\ 0000\end{array}$
(7)	Uranus	$86900000000000000000000000$
(8)	Neptune	$\begin{array}{l}10200000000000000000000\\ 0000\end{array}$
(9)	Pluto	$13100000000000000000000$

We have studied that, a number is written in standard form as a $\times {10}^k$, where a is terminating decimal and k is an integer.

Sl.No.	Name of the Planet	Mass (in kg)
(1)	Mercury	$3.3\times {10}^{23}$
(2)	Venus	$4.87\times {10}^{24}$
(3)	Earth	$5.98\times {10}^{24}$
(4)	Mars	$6.42\times {10}^{23}$
(5)	Jupiter	$1.9\times {10}^{27}$
(6)	Saturn	$5.69\times {10}^{26}$
(7)	Uranus	$8.69\times {10}^{25}$
(8)	Neptune	$1.02\times {10}^{26}$
(9)	Pluto	$1.31\times {10}^{22}$

 

We have studied that, to compare two numbers written in scientific notation: the number with the larger power of $10$ is greater than the number with the smaller power of $10$. If the powers of ten are the same, then the number with larger factor is the larger number.

Hence, the size of the planets arranged in descending order will be Jupiter > Saturn > Neptune > Uranus > Earth > Venus > Mars > Mercury > Pluto

Write the number of seconds in scientific notation.

Sl. No.	Unit	Value in Seconds
(1)	$1$ Minute	$60$
(2)	$1$ Hour	$3,600$
(3)	$1$ Day	$86,400$
(4)	1 Month	$2,600,000$
(5)	1 Year	$32,000,000$
(6)	10 Years	$3,20,000,000$

(1)          We know that, $1$ min $=60s=6\times {10}^{1}s$

(2)          We know that, $1$ hour $=3,600s=36\times {10}^{2}s=3.6\times 10\times {10}^{2}s$ $=3.6\times {10}^{3}s$

(3)          We know that, $1\text{day}=86,400s=8.64\times 100\times {10}^{2}s$
$=8.6\times {10}^{4}s$

(4)          We know that, $1\text{month}=2,600,000s=26\times {10}^{5}s$
$\begin{array}{l}=2.6\times 10\times {10}^{5}s\\ =2.6\times {10}^{6}s\end{array}$

(5)          We know that, $1\text{yr}=32,000,000s=32\times {10}^{6}s$

$=3.2\times 10\times {10}^{6}s=32\times {10}^{7}s$

(6)          We know that, $10\text{yr}=320,000,000s=32\times {10}^{7}s$
$=3.2\times 10\times {10}^{7}s=3.2\times {10}^{8}s$

In our own planet Earth, $361,419,000$ square kilometer of area is covered with water and $148,647,000$ square kilometer of area is covered by land. Find the approximate ratio of area covered with water to area covered by land by converting these numbers into scientific notation.

As per the question, it is given that

Area covered by water $=361419000{\text{km}}^2$

Area covered by land $=148647000{\text{km}}^2$

 In scientific notation, these numbers are written as -

$361419000=361419\times {10}^3$

$\text{Also}, 361419=3.61419\times {10}^5$

So, $$$3.61419\times {10}^5 \times {10}^3 =3.16419\times {10}^8$

$\therefore$  Area covered by water $=3.61419\times {10}^8 {\text{km}}^2$

Similarly,  $148647000=148647\times {10}^3$

Also,   $148647=1.48647\times {10}^5$

So, $1.48647\times {10}^5 \times {10}^3 =1.48647\times {10}^8$

$\therefore$ Area covered by land  $=1.4867\times {10}^8{\text{km}}^2$

 Let $3.61419\times {10}^8 \approx 3.6\times {10}^8$

and   $1.4867\times {10}^8 \approx 1.5\times {10}^8$

Ratio of areas of water to land $=\frac{3.6}{1.5}=12:5$

If $2^{n+2}-2^{n+1}+2^n=c\times 2^n,$ find the value of c.

Considering the given expression, $2^{n+2} -2^{n+1} +2^n=c\times 2^n$

$2^n 2^2 +2^n{2}^1 +2^n =c\times 2^n$             $\left[\therefore a^{m+n}=a^m\times a^n\right]$

$2^n [2^2 –2^1 +1]=c\times 2^n$                 [taking common $2^n$ in LHS]

$2^n [4-2+1]=c\times 2^n$

$3\times 2^n =c\times 2^n$

$3\times 2^n \times 2^{-n} =c\times 2^n \times 2^{-n}$              [multiplying both sides  by $2^n$ ]

$3\times 2^n{}^{-n} =c\times 2^{n-n}$                        $\left[\therefore a^{m+n}=a^m\times a^n\right]$

$3\times 2°=c\times 2°$

$3\times 1=c\times 1$                                     $\left[\therefore a°=1\right]$ 

$\therefore 3=c$

A light year is the distance that light can travel in one year.

1 light year $=9,460,000,000,000$ km.

a.    Express one light year in scientific notation.

b.   The average distance between Earth and Sun is $1.496\times {10}^8$ km. Is the distance between Earth and the Sun greater than, less than or equal  to one light year?

a.    As per question, it is given that 1 light year $=9,460,000,000,000\text{ km}$

for standard form $=946\times {10}^{10}\text{km}$

$=9.46\times {\text{10}}^{\text{2}}\times {10}^{10} \text{km}$

$=9.46\times {10}^{12} \text{km}$

b.   As per question, it is given that  - The average distance between Earth & sun $=1.496\times {10}^8\text{km}$

Distance between Earth & sun $=\frac{1.496}{10000}\times {10}^8 \times {10}^4\text{km}$

$=0.0001496\times {10}^{12}\text{km}$

Since, $9.46>0.0001496$

So, the distance between Earth & sun is less than one light year.

Geometry Application: The number of diagonals of an $n$ -sided figure is $\frac{1}{2}\left(n^2-3n\right)$. Use the formula to find the number of diagonals for a $6$ -sided figure (hexagon).

It is given in the question that, a polygon has $n$ sides, then number of diagonals is $\frac{1}{2}\left(n^2-3n\right)$

In hexagon, there are 6 sides

Therefore, put $n=6$ in the above formula

Number of diagonals $=\frac{1}{2}\left(n^2-3n\right)=\frac{1}{2}\left(6^2-3\left(6\right)\right)$

$=\frac{1}{2}\left(36-18\right)=\frac{1}{2}\times 18=9$

Hence, a hexagon has 9 diagonals.

Life Science: Bacteria can divide in every $20$ minutes. So $1$ bacterium can multiply to $2$ in $20$ minutes. $4$ in $40$ minutes, and so on. How many bacteria will there be in $6$ hours? Write your answer using exponents, and then evaluate.

We already know that, $1\text{h}=60$ min

Therefore, $6h=60\times 6$ min $=360$ min.

As per the question it is given that, a bacterium doubles itself in every $20$ min. Number of times it will double itself $=\frac{360\text{ min}}{20\text{ min}}=18$ 

$\therefore$ Number of bacteria in $6\text{h}=2\times 2\times 2\times \mathrm{...}\times 2\left(18\text{times}\right)=2^{18}$

Blubber makes up $27$ per cent of a blue whale’s body weight. Deepak found the average weight of blue whales and used it to calculate the average weight of their blubber. He wrote the amount as $2^2\times 3^2\times 5\times 17$ kg. Evaluate this amount.

As per the question it is given that, weight calculated by Deepak $=2^2 \times 3^2 \times 5\times 17\text{kg}$

$=2\times 2\times 3\times 3\times 5\times 17=4\times 9\times 5\times 17$

$=36\times 5\times 17=180\times 17=3060\text{kg}$

 Hence, weight calculated by Deepak was $3060$ kg.

Life Science Application: The major components of human blood are red blood cells, white blood cells, platelets and plasma. A typical red blood cell has a diameter of approximately $7\times {10}^{-6}$ metres. A typical platelet has a diameter of approximately $2.33\times {10}^{-6}$ metre. Which has a greater diameter, a red blood cell or a platelet?

As per the question it is given that, diameter of red blood cell $=7\times {10}^{-6}\text{m}$ and diameter of platelet $=2.33\times {10}^{–6}\mathrm{metre}$

We have studied that, two numbers written in scientific notation can be compared. The number with the larger power of $10$ is greater than the number with the smaller power of $10$. If the powers of ten are the same, then the number with the larger factor is the larger number.

Thus, $7\times {10}^{-6}\text{m}$ > $2.33\times {10}^{–6}\mathrm{metre}$

Therefore, red blood cell has a greater diameter than a platelet

A googol is the number $1$ followed by $100$ zeroes.

a.    How is a googol written as a power?

b.   How is a googol times a googol written as a power?

a.    1 googol $=1\times {10}^{100}$

[as there are 100 zeroes after 1]

b.   Googol times googol means

multiply googol by googol

Required number = googol x googol

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

$={10}^{100}{}^{+100}$

$={100}^{200}$

What’s the error?

A student said that $\frac{3^5}{9^5}$ is the same as $\frac{1}{3}$. What mistake has the student made?

According to the question, we have, $\frac{3^5}{9^5}=\frac{3^5}{{\left(3^2\right)}^5}$

$=\frac{3^5}{3^{10}}=\frac{1}{3^{10-5}}=\frac{1}{3^5}$

So, $\frac{1}{3}$ is not same as $\frac{1}{3^5}$

Student has multiplied the base by its exponent. This is an error.