Chapter 13: Exponents and Powers

# Question: 1

Find the value of:

(i)             ${2}^{6}$

(ii)          ${9}^{3}$

(iii)      ${11}^{2}$

(iv)       ${5}^{4}$

## Solution

(i)             ${2}^{6}=2×2×2×2×2×2=64$

(ii)          ${9}^{3}=9×9×9=729$

(iii)      ${11}^{2}=11×11=121$

(iv)       ${5}^{4}=5×5×5×5=625$

# Question: 2

Express the following in exponential form:

(i)             $6×6×6×6$

(ii)          $t×t$

(iii)      $b×b×b×b$

(iv)       $5×5×7×7×7$

(v)          $2×2×a×a$

(vi)       $a×a×a×c×c×c×c×d$

## Solution

(i)             $6×6×6×6={6}^{4}$

(ii)          $t×t={t}^{2}$

(iii)      $b×b×b×b={b}^{4}$

(iv)       $5×5×7×7×7={5}^{2}×{7}^{3}$

(v)          $2×2×a×a={2}^{2}×{a}^{2}$

(vi)       $a×a×a×c×c×c×c×d={a}^{3}×{c}^{4}×d$

# Question: 3

Express each of the following numbers using exponential notation:

(i)             512

(ii)          343

(iii)      729

(iv)       3125

## Solution

(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, $512=2×2×2×2×2×2×2×2×2={2}^{9}$

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

 7 343 7 49 7 7 1

So, $343=7×7×7={7}^{3}$

(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, $729=3×3×3×3×3×3={3}^{6}$

(iv)       3125

We can reduce the number 3125 as given below

 5 3125 5 625 5 125 5 25 5 5 1

Hence, $3125=5×5×5×5×5={5}^{5}$

# Question: 4

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

(i)

(ii)

(iii)

(iv)

(v)

## Solution

(i)             ${4}^{3}=4×4×4=64$

${3}^{4}=3×3×3×3=81$

Since $64<81$

Thus, ${3}^{4}$ is greater than ${4}^{3}$.

(ii)          ${5}^{3}=5×5×5=125$

${3}^{5}=3×3×3×3×3=243$

Since, $125<243$

Thus, ${3}^{5}$ is greater than ${5}^{3}$.

(iii)      ${2}^{8}=2×2×2×2×2×2×2×2=256$

${8}^{2}=8×8=64$

Since, $256>64$

Thus, ${2}^{8}$ is greater than ${8}^{2}$.

(iv)       ${100}^{2}=100×100=10,000$

${2}^{10}=2×2×2×2×2×2×2×2×2×2=1,024$

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

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

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

${10}^{2}=10×10=100$

Since, $1,024>100$

Thus,

# Question: 5

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

(i)             648

(ii)          405

(iii)      540

(iv)       3,600

## Solution

(i)             $648$

Reducing 648 in prime factors

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

$648={2}^{3}×{3}^{4}$

(ii)          $405$

Reducing 405 into its prime factors.

 5 405 3 81 3 27 3 9 3 3 1

$405=5×{3}^{4}$

(iii)      $540$

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

Hence, $540={2}^{2}×{3}^{3}×5$

(iv)       $3,600$

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

$3,600={2}^{4}×{3}^{2}×{5}^{2}$

# Question: 6

Simplify:

(i)             $2×{10}^{3}$

(ii)          ${7}^{2}×{2}^{2}$

(iii)      ${2}^{3}×5$

(iv)       $3×{4}^{4}$

(v)          $0×{10}^{2}$

(vi)       ${5}^{2}×{3}^{3}$

(vii)    ${2}^{4}×{3}^{2}$

(viii)${3}^{2}×{10}^{4}$

## Solution

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

(ii)          ${7}^{2}×{2}^{2}\text{}=7×7×2×2=196$

(iii)      ${2}^{3}×5\text{}=2×2×2×5=40$

(iv)       $3×{4}^{4}\text{}=3×4×4×4×4=768$

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

(vi)       ${5}^{2}×{3}^{3}\text{}=5×5×3×3×3=675$

(vii)    ${2}^{4}×{3}^{2}\text{}=2×2×2×2×3×3=144$

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

# Question: 7

Simplify:

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

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

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

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

## Solution

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

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

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

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

# Question: 8

Compare the following numbers:

(i)

(ii)

## Solution

(i)
On comparing the exponents of base $10$,

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

Therefore,

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

# Question: 1

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

(i)             ${3}^{2}×{3}^{4}×{3}^{8}$

(ii)          ${6}^{15}÷{6}^{10}$

(iii)      ${a}^{3}×{a}^{2}$

(iv)       ${7}^{x}×{7}^{2}$

(v)          ${\left({5}^{2}\right)}^{3}÷{5}^{3}$

(vi)       ${2}^{5}×{5}^{5}$

(vii)    ${a}^{4}×{b}^{4}$

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

(ix)       $\left({2}^{20}÷{2}^{15}\right)×{2}^{3}$

(x)          ${8}^{t}÷{8}^{2}$

## Solution

(i)             ${3}^{2}×{3}^{4}×{3}^{8}={3}^{\left(2+4+8\right)}={3}^{14}$

(ii)          ${6}^{15}÷{6}^{10}={6}^{15-10}={6}^{5}$

(iii)      ${a}^{3}×{a}^{2}={a}^{3+2}={a}^{5}$

(iv)       ${7}^{x}×{7}^{2}={7}^{x+2}$

(v)          $\begin{array}{l}{\left(}^{{5}^{2}}÷{5}^{3}={5}^{2×3}÷{5}^{3}\\ ={5}^{6}÷{5}^{3}\\ ={5}^{6-3}\\ ={5}^{3}\end{array}$

(vi)       ${2}^{5}×{5}^{5}={\left(2×5\right)}^{5}={10}^{5}$

(vii)    ${a}^{4}×{b}^{4}={\left(a×b\right)}^{4}$

(viii) ${\left({\text{3}}^{4}\right)}^{3}={3}^{4×3}={3}^{12}$

(ix)       $\begin{array}{l}\left({2}^{20}÷{2}^{15}\right)×{2}^{3}\\ =\left({2}^{20-15}\right)×{2}^{3}\\ ={2}^{5}×{2}^{3}\\ ={2}^{5+3}\\ ={2}^{8}\end{array}$

(x)          ${8}^{t}÷{8}^{2}={8}^{t-2}$

# Question: 2

Simplify and express each of the following in exponential form:

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

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

(iii)      ${25}^{4}÷{5}^{3}$

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

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

(vi)       ${2}^{0}+{3}^{0}+{4}^{0}$

(vii)    ${2}^{0}×{3}^{0}×{4}^{0}$

(viii)$\left({3}^{0}+{2}^{0}\right)×{5}^{0}$

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

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

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

(xii)    ${\left({2}^{3}×2\right)}^{2}$

## Solution

(i)

(ii)          $\begin{array}{l}{25}^{4}÷{5}^{3}\\ ={\left(}^{{5}^{2}}÷{5}^{3}\\ ={5}^{8}÷{5}^{3}\end{array}$
$\begin{array}{l}={5}^{8-3}\\ ={5}^{5}\end{array}$

(iii)      $\begin{array}{l}\frac{3×{7}^{2}×{11}^{8}}{21×{11}^{3}}\\ =\frac{3×{7}^{2}×{11}^{8}}{3×7×{11}^{3}}\end{array}$

(iv)

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

(vi)       $\begin{array}{l}{2}^{0}×{3}^{0}×{4}^{0}\\ =1×1×1\\ =1\end{array}$

(vii)    $\begin{array}{l}\left({3}^{0}+{2}^{0}\right)×{5}^{0}\\ =\left(1+1\right)×1\\ =2×1\\ =2\end{array}$

(viii)$\begin{array}{l}\frac{{2}^{8}×{a}^{5}}{{4}^{3}×{a}^{3}}\\ =\frac{{2}^{8}×{a}^{5}}{{\left({2}^{2}\right)}^{3}×{a}^{3}}\end{array}$

(ix)       $\begin{array}{l}\left(\frac{{a}^{5}}{{a}^{3}}\right)×{a}^{8}\\ =\left({a}^{5-3}\right)×{a}^{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\because {a}^{m}÷{a}^{n}={a}^{m-n}\right]\\ ={a}^{2}×{a}^{8}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ ={a}^{2+8}={a}^{10}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\because {a}^{m}×{a}^{n}={a}^{m+n}\right]\end{array}$

(x)          $\begin{array}{l}\frac{{4}^{5}×{a}^{8}{b}^{3}}{{4}^{5}×{a}^{5}{b}^{2}}\\ ={4}^{5-5}×{a}^{8-5}×{b}^{3-2}\\ ={4}^{0}×{a}^{3}×b\\ =1×{a}^{3}×b\\ ={a}^{3}b\end{array}$                         . .

(xi)       ${\left({2}^{3}×2\right)}^{2}$

# Question: 3

Say true or false and justify your answer:

(i)             $10×{10}^{11}={100}^{11}$

(ii)          ${2}^{3}>{5}^{2}$

(iii)      ${2}^{3}×{3}^{2}={6}^{5}$

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

## Solution

(i)             $10×{10}^{11}={100}^{11}$

and

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

Therefore, it is false.

(ii)          ${2}^{3}>{5}^{2}$

and

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

Therefore, it is false.

(iii)      ${2}^{3}×{3}^{2}={6}^{5}$

and

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

Therefore, it is false.

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

and

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

Therefore, it is true.

# Question: 4

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

(i)             $108×192$

(ii)          $270$

(iii)      $729×64$

(iv)       $768$

## Solution

(i)             $108×192$

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

 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, $270=2×{3}^{3}×5$

(iii)      $729×64$

 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, $729×64={3}^{6}×{2}^{6}$

(iv)       $768$

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

Hence, $768={2}^{8}×3$

# Question: 5

Simplify:

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

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

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

## Solution

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

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

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

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

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

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

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

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

# Question: 1

Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

## Solution

(i)             $2,79,404$

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

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

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

(ii)          $30,06,194$

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

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

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

(iii)      $28,06,196$

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

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

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

(iv)       $1,20,719$

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

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

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

(v)          $20,068$

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

$=2×10000+0×1000+0×100+6×10+8×1$

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

# Question: 2

Find the number from each of the following expanded forms:

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

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

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

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

## Solution

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

$=8×10000+6×1000+0×100+4×10+5×1$

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

$=86,045$

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

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

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

$=4,05,302$

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

$=3×10000+0×1000+7×100+0×10+5×1$

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

$=30,705$

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

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

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

$=9,00,230$

# Question: 3

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

## Solution

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

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

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

$=3.1865×10000×100000=3.1865×{10}^{9}$

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

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

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

# Question: 4

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

a.    The distance between Earth and Moon is .

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.

## Solution

a.    The distance between Earth and Moon

$=3.84×100×1000000$

b.   Speed of light in vacuum

c.    Diameter of the Earth

d.   Diameter of the Sun

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

f.     Years of Universe

g.   Distance of the Sun from the centre of the Milky Way Galaxy

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

i.      The Earth has Sea water

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