10.1 Introduction

Do you know?

Mass of earth is 5,970,000,000,000, 000, 000, 000, $000kg$. We have already learnt in earlier class how to write such large numbers more conveniently using exponents, as, $5.97\times {10}^{24}kg$.

We read ${10}^{24}$ as 10 raised to the power 24 .

We know $$ $$ $2^5=2\times 2\times 2\times 2\times 2$

and $$ $$ $2^m=2\times 2\times 2\times 2\times \dots \times 2\times 2\dots (m\text{ times })$

Let us now find what is $2^{-2}$ is equal to?

10.2 Powers with Negative Exponents

You know that,

$\begin{array}{rl}& {10}^1=10=\frac{100}{10}\\ & {10}^1=10=\frac{100}{10}\\ & {10}^0=1=\frac{10}{10}\\ & {10}^{-1}=?\end{array}$

As the exponent decreases by1, the value becomes one-tenth of the previous value.

Continuing the above pattern we get, ${10}^{-1}=\frac{1}{10}$

Similarly

$\begin{array}{rl}& {10}^{-2}=\frac{1}{10}÷10=\frac{1}{10}\times \frac{1}{10}=\frac{1}{100}=\frac{1}{{10}^2}\\ & {10}^{-3}=\frac{1}{100}÷10=\frac{1}{100}\times \frac{1}{10}=\frac{1}{1000}=\frac{1}{{10}^3}\end{array}$

What is ${10}^{-10}$ equal to?

Now consider the following.

So looking at the above pattern, we say

$\begin{array}{rl}& 3^{-1}=1÷3=\frac{1}{3}\\ & 3^{-2}=\frac{1}{3}÷3=\frac{1}{3\times 3}=\frac{1}{3^2}\\ & 3^{-3}=\frac{1}{3^2}÷3=\frac{1}{3^2}\times \frac{1}{3}=\frac{1}{3^3}\end{array}$

You can now find the value of $2^{-2}$ in a similar manner.

We have,

$\begin{array}{ccccc}{10}^{-2}& =\frac{1}{{10}^2}& \text{ or }& {10}^2& =\frac{1}{{10}^{-2}}\\ {10}^{-3}& =\frac{1}{{10}^3}& \text{ or }& {10}^3& =\frac{1}{{10}^{-3}}\\ 3^{-2}& =\frac{1}{3^2}& \text{ or }& 3^2& =\frac{1}{3^{-2}}& \text{ etc. }\end{array}$

In general, we can say that for any non-zero integer $a,a^{-m}=\frac{1}{a^m}$, where $m$ is a positive integer. $a^{-m}$ is the multiplicative inverse of $a^m$.

TRY THESE

Find the multiplicative inverse of the following.

(i) $2^{-4}$ $$ (ii) ${10}^{-5}$ $$ (iii) $7^{-2}$ $$ (iv) $5^{-3}$ $$ (v) ${10}^{-100}$

We learnt how to write numbers like 1425 in expanded form using exponents as $1\times {10}^3+4\times {10}^2+2\times {10}^1+5\times {10}^{\circ }$.

Let us see how to express 1425.36 in expanded form in a similar way.

We have $1425.36=1\times 1000+4\times 100+2\times 10+5\times 1+\frac{3}{10}+\frac{6}{100}$

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

${10}^{-1}=\frac{1}{10},{10}^{-2}=\frac{1}{{10}^2}=\frac{1}{100}$

TRY THESE

Expand the following numbers using exponents.

(i) 1025.63 $$ (ii) 1256.249

10.3 Laws of Exponents

We have learnt that for any non-zero integer $a,a^m\times a^n=a^{m+n}$, where $m$ and $n$ are natural numbers. Does this law also hold if the exponents are negative? Let us explore.

(i)

Therefore, $2^{-3}\times 2^{-2}=\frac{1}{2^3}\times \frac{1}{2^2}=\frac{1}{2^3\times 2^2}=\frac{1}{2^{3+2}}=2^{-}$

(ii) Take $(-3{)}^{-4}\times (-3{)}^{-3}$

$\begin{array}{rl}(-3{)}^{-4}\times (-3{)}^{-3}& =\frac{1}{(-3{)}^4}\times \frac{1}{(-3{)}^3}\\ & =\frac{1}{(-3{)}^4\times (-3{)}^3}=\frac{1}{(-3{)}^{4+3}}=(-3{)}^{-7}\end{array}$

(iii) Now consider $5^{-2}\times 5^4$

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

In Class VII, you have learnt that for any non-zero integer $a,\frac{a^m}{a^n}=a^{m-n}$, where

(iv) Now consider $(-5{)}^{-4}\times (-5{)}^2$ $m$ and $n$ are natural numbers and $m>n$.

$\begin{array}{rl}(-5{)}^{-4}\times (-5{)}^2& =\frac{1}{(-5{)}^4}\times (-5{)}^2=\frac{(-5{)}^2}{(-5{)}^4}=\frac{1}{(-5{)}^4\times (-5{)}^{-2}}\\ & =\frac{1}{(-5{)}^{4-2}}=(-5{)}^{-(2)}⟶\underbrace{(-4)+2=-2}\end{array}$

In general, we can say that for any non-zero integer $a$, $a^m\times a^n=a^{m+n}$, where $m$ and $n$ are integers.

TRY THESE

Simplify and write in exponential form.

(i) $(-2{)}^{-3}\times (-2{)}^{-4}$ $$ (ii) $p^3\times p^{-10}$ $$ (iii) $3^2\times 3^{-5}\times 3^6$

On the same lines you can verify the following laws of exponents, where $a$ and $b$ are non zero integers and $m,n$ are any integers.

(i) $\frac{a^m}{a^n}=a^{m-n}$ $$ (ii) $(a^m{)}^n=a^{mn}$ $$ (iii) $a^m\times b^m=(ab{)}^m$

(iv) $\frac{a^m}{b^m}=(\frac{a}{b}{)}^m$ $$ (v) $a^0=1$

Let us solve some examples using the above Laws of Exponents.

Example 1 : Find the value of (i) $2^{-3}$ (ii) $\frac{1}{3^{-2}}$

Solution:

(i) $2^{-3}=\frac{1}{2^3}=\frac{1}{8}$ (ii) $\frac{1}{3^{-2}}=3^2=3\times 3=9$

Example 2 : Simplify

(i) $(-4{)}^5\times (-4{)}^{-10}$ (ii) $2^5÷2^{-6}$ $$

Solution:

(i) $(-4{)}^5\times (-4{)}^{-10}=(-4{)}^{(5-10)}=(-4{)}^{-5}=\frac{1}{(-4{)}^5}(a^m\times a^n=a^{m+n},a^{-m}=\frac{1}{a^m})$

(ii) $2^5÷2^{-6}=2^{5-(-6)}=2^{11}(a^m÷a^n=a^{m-n})$

Example 3 : Express $4^{-3}$ as a power with the base 2.

Solution: We have, $4=2\times 2=2^2$

Therefore, $(4{)}^{-3}=(2\times 2{)}^{-3}=(2^2{)}^{-3}=2^{2\times (-3)}=2^{-6}[(a^m{)}^n=a^{mn}]$

Example 4 : Simplify and write the answer in the exponential form.

(i) $(2^5÷2^8{)}^5\times 2^{-5}$

(iii) $\frac{1}{8}\times (3{)}^{-3}$ (ii) $(-4{)}^{-3}\times (5{)}^{-3}\times (-5{)}^{-3}$

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

Solution:

(i) $(2^5÷2^8{)}^5\times 2^{-5}=(2^{5-8}{)}^5\times 2^{-5}=(2^{-3}{)}^5\times 2^{-5}=2^{-15-5}=2^{-20}=\frac{1}{2^{20}}$

(ii) $(-4{)}^{-3}\times (5{)}^{-3}\times (-5{)}^{-3}=[(-4)\times 5\times (-5){]}^{-3}=[100{]}^{-3}=\frac{1}{{100}^3}$

[using the law $a^m\times b^m=(ab{)}^m,a^{-m}=\frac{1}{a^m}$ ]

(iii) $\frac{1}{8}\times (3{)}^{-3}=\frac{1}{2^3}\times (3{)}^{-3}=2^{-3}\times 3^{-3}=(2\times 3{)}^{-3}=6^{-3}=\frac{1}{6^3}$

(iv) $(-3{)}^4\times (\frac{5}{3}{)}^4=(-1\times 3{)}^4\times \frac{5^4}{3^4}=(-1{)}^4\times 3^4\times \frac{5^4}{3^4}$

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

Example 5 : Find $m$ so that $(-3{)}^{m+1}\times (-3{)}^5=(-3{)}^7$

Solution: $(-3{)}^{m+1}\times (-3{)}^5=(-3{)}^7$

$\begin{array}{rl}(-3{)}^{m+1+5}& =(-3{)}^7\\ (-3{)}^{m+6}& =(-3{)}^7\end{array}$

On both the sides powers have the same base different from 1 and -1 , so their exponents must be equal.

Therefore, $m+6=7$

or $m=7-6=1$

Example 6 : Find the value of $(\frac{2}{3}{)}^{-2}$.

$a^n=1$ only if $n=0$. This will work for any $a$. For $a=1,1^1=1^2=1^3=1^{-2}=\dots =1$ or $(1{)}^n=$ 1 for infinitely many $n$.

For $a=-1$,

$(-1{)}^0=(-1{)}^2=(-1{)}^4=(-1{)}^{-2}=\dots =1$ or $(-1{)}^p=1$ for any even integer $p$.

Solution: $(\frac{2}{3}{)}^{-2}=\frac{2^{-2}}{3^{-2}}=\frac{3^2}{2^2}=\frac{9}{4}$

Example 7 : Simplify (i) $(\frac{1}{3}{)}^{-2}-(\frac{1}{2}{)}^{-3}÷(\frac{1}{4}{)}^{-2}$

Solution:

(i) $(\frac{1}{3}{)}^{-2}-(\frac{1}{2}{)}^{-3}÷(\frac{1}{4}{)}^{-2}=\frac{1^{-2}}{3^{-2}}-\frac{1^{-3}}{2^{-3}}÷\frac{1^{-2}}{4^{-2}}$

$=\frac{3^2}{1^2}-\frac{2^3}{1^3}÷\frac{4^2}{1^2}=9-8÷16=\frac{1}{16}$

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

$=5^{-2}\times 8^2=\frac{8^2}{5^2}=\frac{64}{25}$

EXERCISE 10.1

1. Evaluate.

(i) $3^{-2}$

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

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

2. Simplify and express the result in power notation with positive exponent.

(i) $(-4{)}^5÷(-4{)}^8$

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

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

(iv) $(3^{-7}÷3^{-10})\times 3^{-5}$

(v) $2^{-3}\times (-7{)}^{-3}$

3. Find the value of

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

(ii) $(2^{-1}\times 4^{-1})÷2^{-2}$

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

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

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

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

6. Evaluate (i) ${(\frac{1}{3}{)}^{-1}-(\frac{1}{4}{)}^{-1}}^{-1}$ (ii) $(\frac{5}{8}{)}^{-7}\times (\frac{8}{5}{)}^{-4}$

7. Simplify. (i) $\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}(t\ne 0)$ (ii) $\frac{3^{-5}\times {10}^{-5}\times 125}{5^{-7}\times 6^{-5}}$

10.4 Use of Exponents to Express Small Numbers in Standard Form

Observe the following facts.

1. The distance from the Earth to the Sun is $149,600,000,000m$.

2. The speed of light is $300,000,000m/sec$.

3. Thickness of Class VII Mathematics book is $20mm$.

4. The average diameter of a Red Blood Cell is $0.000007mm$.

5. The thickness of human hair is in the range of $0.005cm$ to $0.01cm$.

6. The distance of moon from the Earth is $384,467,000m$ (approx).

7. The size of a plant cell is $0.00001275m$.

8. Average radius of the Sun is $695000km$.

9. Mass of propellant in a space shuttle solid rocket booster is $503600kg$.

10. Thickness of a piece of paper is $0.0016cm$.

11. Diameter of a wire on a computer chip is $0.000003m$.

12. The height of Mount Everest is $8848m$.

Observe that there are few numbers which we can read like $2cm,8848m$, $6,95,000km$. There are some large numbers like $150,000,000,000m$ and some very small numbers like $0.000007m$.

Identify very large and very small numbers from the above facts and write them in the adjacent table:

We have learnt how to express very large numbers in standard form in the previous class.

For example: $150,000,000,000=1.5\times {10}^{11}$

Now, let us try to express $0.000007m$ in standard form.

$\begin{array}{rl}0.000007& =\frac{7}{1000000}=\frac{7}{{10}^6}=7\times {10}^{-6}\\ 0.000007m& =7\times {10}^{-6}m\end{array}$

Similarly, consider the thickness of a piece of paper which is $0.0016cm$.

$\begin{array}{rl}0.0016& =\frac{16}{10000}=\frac{1.6\times 10}{{10}^4}=1.6\times 10\times {10}^{-4}\\ & =1.6\times {10}^{-3}\end{array}$

Therefore, we can say thickness of paper is $1.6\times {10}^{-3}cm$.

Again notice

0.0016 decimal is moved 1233 places to the right.

TRY THESE

1. Write the following numbers in standard form.

(i) 0.000000564

(ii) 0.0000021

(iii) 21600000

(iv) 15240000

2. Write all the facts given in the standard form.

10.4.1 Comparing very large and very small numbers

The diameter of the Sun is $1.4\times {10}^{9}m$ and the diameter of the Earth is $1.2756\times {10}^{7}m$.

Suppose you want to compare the diameter of the Earth, with the diameter of the Sun.

Diameter of the Sun $=1.4\times {10}^{9}m$

Diameter of the earth $=1.2756\times {10}^{7}m$

Therefore $\frac{1.4\times {10}^9}{1.2756\times {10}^7}=\frac{1.4\times {10}^{9-7}}{1.2756}=\frac{1.4\times 100}{1.2756}$ which is approximately 100

So, the diameter of the Sun is about 100 times the diameter of the earth.

Let us compare the size of a Red Blood cell which is $0.000007m$ to that of a plant cell which is $0.00001275m$.

$\begin{array}{rl}& \text{ Size of Red Blood cell }=0.000007m=7\times {10}^{-6}m\\ & \text{ Size of plant cell }=0.00001275=1.275\times {10}^{-5}m\end{array}$

Therefore, $\frac{7\times {10}^{-6}}{1.275\times {10}^{-5}}=\frac{7\times {10}^{-6-(-5)}}{1.275}=\frac{7\times {10}^{-1}}{1.275}=\frac{0.7}{1.275}=\frac{0.7}{1.3}=\frac{1}{2}$ (approx.)

So a red blood cell is half of plant cell in size.

Mass of earth is $5.97\times {10}^{24}kg$ and mass of moon is $7.35\times {10}^{22}kg$. What is the total mass?

$\begin{array}{rl}\text{ Total mass }& =5.97\times {10}^{24}kg+7.35\times {10}^{22}kg.\\ & =5.97\times 100\times {10}^{22}+7.35\times {10}^{22}\\ & =597\times {10}^{22}+7.35\times {10}^{22}\\ & =(597+7.35)\times {10}^{22}\\ & =604.35\times {10}^{22}kg.\end{array}$

$\boxed{{\displaystyle \text{When we have to add numbers i standard form, we convert them into numbers with the same exponents}}}$

The distance between Sun and Earth is $1.496\times {10}^{11}m$ and the distance between

Earth and Moon is $3.84\times {10}^{8}m$.

During solar eclipse moon comes in between Earth and Sun.

At that time what is the distance between Moon and Sun.

$\begin{array}{rl}\text{ Distance between Sun and Earth }& =1.496\times {10}^{11}\mathrm{m}\\ \text{ Distance between Earth and Moon }& =3.84\times {10}^8\mathrm{m}\\ \text{ Distance between Sun and Moon }& =1.496\times {10}^{11}-3.84\times {10}^8\\ & =1.496\times 1000\times {10}^8-3.84\times {10}^8\\ & =(1496-3.84)\times {10}^8\mathrm{m}=1492.16\times {10}^8\mathrm{m}\end{array}$

Example 8 : Express the following numbers in standard form. (i) 0.000035 (ii) 4050000 Solution (i) $0.000035=3.5\times {10}^{-5}$ (ii) $4050000=4.05\times {10}^6$

Example 9 : Express the following numbers in usual form. (i) $3.52\times {10}^5$ (ii) $7.54\times {10}^{-4}$ (iii) $3\times {10}^{-5}$

Solution:

EXERCISE 10.2

1. Express the following numbers in standard form.

(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000

2. Express 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) $3.61492\times {10}^6$

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

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

(ii) Charge of an electron is $0.000,000,000,000,000,000,16$ coulomb.

(iii) Size of a bacteria is $0.0000005m$

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

(v) Thickness of a thick paper is $0.07mm$

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

WHAT HAVE WE DISCUSSED??

1. Numbers with negative exponents obey the following laws of exponents.

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

(b) $a^m÷a^n=a^{m-n}$

(c) $(a^m{)}^n=a^{mn}$

(d) $a^m\times b^m=(ab{)}^m$

(e) $a^0=1$

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

2. Very small numbers can be expressed in standard form using negative exponents.