Exponents and Powers

INTRODUCTION

Do you know what the mass of earth is? It is 5,970,000,000,000,000,000,000,000 kg!

Can you read this number?

Mass of Uranus is 86,800,000,000,000,000,000,000,000 kg.

Which has greater mass, Earth or Uranus?

Distance between Sun and Saturn is 1,433,500,000,000 m and distance between Saturn and Uranus is 1,439,000,000,000 m. Can you read these numbers? Which distance is less?

      These very large numbers are difficult to read, understand and compare. To make these numbers easy to read, understand and compare, we use exponents. In this Chapter, we shall learn about exponents and also learn how to use them.

Exponents and Powers

EXPONENTS

We can write large numbers in a shorter form using exponents.

Observe 10, 000 = 10 × 10 × 10 × 10 = 10⁴

       The short notation 10⁴ stands for the product 10×10×10×10. Here ‘10’ is called the base and ‘4’ the exponent. The number 10⁴ is read as 10 raised to the power of 4 or simply as fourth power of 10. 10⁴ is called the exponential form of 10,000.

           We can similarly express 1,000 as a power of 10. Note that

                               1000 = 10 × 10 × 10 = 10³

 Here again, 10³ is the exponential form of 1,000.

Similarly, 1,00,000 = 10 × 10 × 10 × 10 × 10 = 10⁵

10⁵ is the exponential form of 1,00,000

In both these examples, the base is 10; in case of 10³, the exponent is 3 and in case of 10⁵ the exponent is 5.

We have used numbers like 10, 100, 1000 etc., while writing numbers in an expanded form. For example, 47561 = 4 × 10000 + 7 × 1000 + 5 × 100 + 6 × 10 + 1

This can be written as 4 × 10⁴ + 7 ×10³ + 5 × 10² + 6 × 10 + 1.

Try writing these numbers in the same way 172, 5642, 6374.

        In all the above given examples, we have seen numbers whose base is 10. However the base can be any other number also. For example:

81 = 3 × 3 × 3 × 3 can be written as 81 = 3⁴ , here 3 is the base and 4 is the exponent.

Some powers have special names. For example,

10² , which is 10 raised to the power 2, also read as ‘10 squared’ and 10³ , which is 10 raised to the power 3, also read as ‘10 cubed’.

Can you tell what 5³ (5 cubed) means?

5 ³ = 5 × 5 × 5 = 125

So, we can say 125 is the third power of 5.

What is the exponent and the base in 5³?

Similarly, 2⁵ = 2 × 2 × 2 × 2 × 2 = 32, which is the fifth power of 2.

In 2⁵ , 2 is the base and 5 is the exponent.

In the same way, 243 = 3 × 3 × 3 × 3 × 3 = 35

                         64 = 2 × 2 × 2 × 2 × 2 × 2 = 26

                         625 = 5 × 5 × 5 × 5 = 54

You can also extend this way of writing when the base is a negative integer. What does (–2)³ mean?

It is (–2)³ = (–2) × (–2) × (–2) = – 8

Is (–2)⁴ = 16? Check it.

Instead of taking a fixed number let us take any integer a as the base, and write the numbers as,

a × a = a ² (read as ‘a squared’ or ‘a raised to the power 2’)

a × a × a = a ³ (read as ‘a cubed’ or ‘a raised to the power 3’)

a × a × a × a = a ⁴ (reaa × a × b × b × b × b can be expressed as a ²b ⁴ (read as a squared into b raised to the power of 4)d as a raised to the power 4 or the 4th power of a) ..............................

a × a × a × a × a × a × a = a ⁷ (read as a raised to the power 7 or the 7th power of a) and so on.

a × a × a × b × b can be expressed as a ³b ² (read as a cubed b squared).

EXAMPLE : Express 256 as a power 2.

SOLUTION We have 256 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2.

So we can say that 256 = 2⁸

EXAMPLE :Which one is greater 2³ or 3² ?

SOLUTION We have, 2³ = 2 × 2 × 2 = 8 and 3 ² = 3 × 3 = 9.

Since 9 > 8, so, 3² is greater than 2³

EXAMPLE :Which one is greater 8² or 2⁸ ?

SOLUTION 8 ² = 8 × 8 = 64

2 ⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256 Clearly, 2 ⁸ > 8 ²

EXAMPLE: Expand a ³ b ² , a ² b ³ , b ² a ³ , b ³ a ² . Are they all same?

SOLUTION

a ³ b ² = a ³ × b ² = (a × a × a) × (b × b) = a × a × a × b × b

a ² b ³ = a ² × b ³ = a × a × b × b × b

b ² a ³ = b ² × a³ = b × b × a × a × a

b ³ a ² = b ³ × a ² = b × b × b × a × a

        Note that in the case of terms a ³ b ² and a ² b ³ the powers of a and b are different. Thus a ³ b ² and a ² b ³ are different.

        On the other hand, a ³ b ² and b ² a ³ are the same, since the powers of a and b in these two terms are the same. The order of factors does not matter. Thus, a ³ b ² = a ³ × b ² = b ² × a ³ = b ² a ³ . Similarly, a ² b ³ and b ³ a ² are the same.

EXAMPLE Express the following numbers as a product of powers of prime factors: (i) 72 (ii) 432 (iii) 1000 (iv) 16000

SOLUTION (i) 72 = 2 × 36

                      = 2 × 2 × 18

                     = 2 × 2 × 2 × 9

                     = 2 × 2 × 2 × 3 × 3

                     = 2³ × 3²

Thus, 72 = 2 ³ × 3² (required prime factor product form)

(ii) 432 = 2 × 216

           = 2 × 2 × 108

           = 2 × 2 × 2 × 54

           = 2 × 2 × 2 × 2 × 27

           = 2 × 2 × 2 × 2 × 3 × 9

           = 2 × 2 × 2 × 2 × 3 × 3 × 3

 or 432 = 2 ⁴ × 3³ (required form)

(iii) 1000 = 2 × 500

              = 2 × 2 × 250

              = 2 × 2 × 2 × 125

               = 2 × 2 × 2 × 5 × 25

               = 2 × 2 × 2 × 5 × 5 × 5 or 1000

  or 1000  = 2 ³ × 5³

Atul wants to solve this example in another way:

1000 = 10 × 100 = 10 × 10 × 10

          = (2 × 5) × (2 × 5) × (2 × 5) (Since10 = 2 × 5)

         = 2 × 5 × 2 × 5 × 2 × 5 = 2 × 2 × 2 × 5 × 5 × 5

or 1000 = 2³ × 5³ Is Atul’s method correct?

(iv) 16,000 = 16 × 1000 = (2 × 2 × 2 × 2) ×1000 = 2⁴ ×10³ (as 16 = 2 × 2 × 2 × 2)

= (2 × 2 × 2 × 2) × (2 × 2 × 2 × 5 × 5 × 5) = 2⁴ × 2³ × 5³

                        (Since 1000 = 2 × 2 × 2 × 5 × 5 × 5)

= (2 × 2 × 2 × 2 × 2 × 2 × 2 ) × (5 × 5 × 5) or, 16,000 = 2 ⁷ × 5³

EXAMPLE 6 Work out (1)⁵ , (–1)³ , (–1)⁴ , (–10)³ , (–5)⁴ .

SOLUTION (i) We have (1)⁵ = 1 × 1 × 1 × 1 × 1 = 1 In fact, you will realise that 1 raised to any power is 1.

(ii) (–1)³ = (–1) × (–1) × (–1) = 1 × (–1) = –1

(iii) (–1)⁴ = (–1) × (–1) × (–1) × (–1) = 1 ×1 = 1

You may check that (–1) raised to any odd power is (–1), and (–1) raised to any even power is (+1).

(iv) (–10)³ = (–10) × (–10) × (–10) = 100 × (–10) = – 1000

(v) (–5)⁴ = (–5) × (–5) × (–5) × (–5) = 25 × 25 = 625

Exponents and Powers

LAWS OF EXPONENTS

Multiplying Powers with the Same Base:

(i) Let us calculate 2² × 2³

2 ² × 2³ = (2 × 2) × (2 × 2 × 2)

              = 2 × 2 × 2 × 2 × 2 = 2⁵ = 2²⁺³

Note that the base in 2² and 2³ is same and the sum of the exponents, i.e., 2 and 3 is 5

(ii) (–3)⁴ × (–3)³ = [(–3) × (–3) × (–3)× (–3)] × [(–3) × (–3) × (–3)] = (–3) × (–3) × (–3) × (–3) × (–3) × (–3) × (–3) = (–3)⁷ = (–3)⁴⁺³ Again, note that the base is same and the sum of exponents, i.e., 4 and 3, is 7

(iii) a ² × a ⁴ = (a × a) × (a × a × a × a) = a × a × a × a × a × a = a ⁶

(Note: the base is the same and the sum of the exponents is 2 + 4 = 6)

Similarly, verify: 4 ² × 4² = 4 ²⁺²

                       3 ² × 3³ = 3 ²⁺³

Can you write the appropriate number in the box.

(–11)² × (–11)⁶ = (–11)^?

b ² × b ³ =b^?  (Remember, base is same; b is any integer).

c ³ × c ⁴ =c^? (c is any integer)

d ¹⁰ × d ²⁰ = d^?

From this we can generalise that for any non-zero integer a, where m and n are whole numbers,

                                       a ^m × a ⁿ = a ^{m + n}

Caution!

Consider 2³ × 3²

Can you add the exponents? No! Do you see ‘why’? The base of 2³ is 2 and base of 3² is 3. The bases are not same.

Exponents and Powers

Dividing Powers with the Same Base

Let us simplify 3⁷ ÷ 3⁴ ?

\({{3}^{7}}\div {{3}^{4}}=\frac{{{3}^{7}}}{{{3}^{4}}}=\frac{3\text{ }\times 3\times 3\times 3\times 3\times 3\times 3}{3\text{ }\times 3\times 3\times 3}\)

= 3 × 3 × 3 = 3³ = 3 ^{7 – 4}

Thus 3 ⁷ ÷ 3⁴ = 3^{7 – 4} (Note, in 3⁷ and 3⁴ the base is same and 3⁷ ÷ 3⁴ becomes 3^7–4 )

Similarly,

                   \({{5}^{6}}\div {{5}^{2}}=\frac{{{5}^{6}}}{{{5}^{2}}}=\frac{5\times 5\times 5\times 5\times 5\times 5}{5\times 5}\)

                                 = 5 × 5 × 5 × 5 = 5⁴ = 5^{6 – 2}

                 or 5 ⁶ ÷ 5² = 5 ^{6 – 2}

Let a be a non-zero integer, then,

            \({{a}^{4}}\div {{a}^{2}}=\frac{{{a}^{4}}}{{{a}^{2}}}=\frac{a\times a\times a\times a}{a\times a}=a\times a={{a}^{2}}={{a}^{4-2}}\)

or a ⁴ ÷ a ² = a ^{4 – 2}

Now can you answer quickly?

10⁸ ÷ 10³ = 10^{8 – 3} = 10⁵

7 ⁹ ÷ 7⁶ = 7^?

a ⁸ ÷ a ⁵ = a^?

For non-zero integers b and c,

b ¹⁰ ÷ b ⁵ = b^?

c ¹⁰⁰ ÷ c ⁹⁰ = c^?

In general, for any non-zero integer a, a ^m ÷ a ⁿ = a ^{m – n} where m and n are whole numbers and m > n.

Exponents and Powers

Taking Power of a Power

Consider the following

Simplify \({{({{2}^{3}})}^{2}};{{({{3}^{2}})}^{4}}\)

Now, \({{({{2}^{3}})}^{2}}\) means 2³ is multiplied two times with itself.

\({{({{2}^{3}})}^{2}}\) = 2³ × 2³

       = 2^{3 + 3} (Since a m × a n = a m + n)

       = 2⁶ = 2^{3 × 2}

Thus \({{({{2}^{3}})}^{2}}\)= 2^3×2

Similarly,      \(\begin{align}   & {{({{3}^{2}})}^{4}}={{3}^{2}}\times {{3}^{2}}\times {{3}^{2}}\times {{3}^{2}} \\  & ={{3}^{2+2+2+2}} \\  & ={{3}^{8}} \\  & ={{3}^{2\times 4}} \\ \end{align}\)

Can you tell what would \({{({{7}^{2}})}^{10}}\) would be equal to?

So \({{({{2}^{3}})}^{2}}\) = 2^{3 × 2} = 2⁶

\({{({{3}^{2}})}^{4}}\) = 3^{2 × 4} = 3⁸ = 7^{2 × 10} = 7²⁰

\({{({{a}^{2}})}^{3}}\) = a ^{2 × 3} = a ⁶ = a ^{m × 3} = a ^3m

From this we can generalise for any non-zero integer ‘a’, where ‘m’ and ‘n’ are whole numbers, \({{({{a}^{m}})}^{n}}\)= a ^mn

EXAMPLE  Can you tell which one is greater (5² ) × 3 or \({{({{5}^{2}})}^{3}}\)?

SOLUTION (5² ) × 3 means 5² is multiplied by 3 i.e., 5 × 5 × 3 = 75

but \({{({{5}^{2}})}^{3}}\) means 5² is multiplied by itself three times i.e. , 5 ² × 5² × 5² = 5 ⁶ = 15,625

Therefore (5² ) ³ > (5² ) × 3

Exponents and Powers

Multiplying Powers with the Same Exponents

Can you simplify 2³ × 3³ ?

Notice that here the two terms 2³ and 3³ have different bases, but the same exponents.

Now, 2 ³ × 3³ = (2 × 2 × 2) × (3 × 3 × 3)

                    = (2 × 3) × (2 × 3) × (2 × 3)

                    = 6 × 6 × 6 = 6³ (Observe 6 is the product of bases 2 and 3)

Consider 4⁴ × 3⁴ = (4 × 4 × 4 × 4) × (3 × 3 × 3 × 3)

                        = (4 × 3) × (4 × 3) × (4 × 3) × (4 × 3)

                        = 12 × 12 × 12 × 12 = 12⁴

Consider, also, 3² × a ² = (3 × 3) × (a × a)

                                 = (3 × a) × (3 × a)

                                    = (3 × a) ² = (3a) ² (Note: 3×a = 3a )

Similarly, a ⁴ × b ⁴ = (a × a × a × a) × (b × b × b × b)

                           = (a × b) × (a × b) × (a × b) × (a × b)

                           = (a × b) ⁴ = (ab) ⁴ (Note a × b = ab)

In general, for any non-zero integer a

a ^m × b ^m = (ab) ^m (where m is any whole number)

EXAMPLE 8Express the following terms in the exponential form: (i) (2 × 3)⁵ (ii) (2a) ⁴ (iii) (– 4m) ³

SOLUTION

(i) (2 × 3)⁵ = (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3) × (2 × 3)

                = (2 × 2 × 2 × 2 × 2) × (3 × 3× 3 × 3 × 3)

                = 2⁵ × 3⁵

(ii) (2a) ⁴ = 2a × 2a × 2a × 2a

              = (2 × 2 × 2 × 2) × (a × a × a × a)

              = 2⁴ × a ⁴

(iii) (– 4m) ³ = (– 4 × m) ³

                  = (– 4 × m) × (– 4 × m) × (– 4 × m)

                  = (– 4) × (– 4) × (– 4) × (m × m × m)

                  = (– 4)³ × (m) ³

Exponents and Powers

Dividing Powers with the Same Exponents

Observe the following simplifications:

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

(ii) \(\frac{{{a}^{3}}}{{{b}^{3}}}=\frac{a\times a\times a}{b\times b\times b}=\frac{a}{b}\times \frac{a}{b}\times \frac{a}{b}={{\left( \frac{a}{b} \right)}^{3}}\)

From these examples we may generalise

\({{a}^{m}}\div {{b}^{m}}=\frac{{{a}^{m}}}{{{b}^{m}}}={{\left( \frac{a}{b} \right)}^{m}}\) where a and b are any non zero integers and m is a whole number.

EXAMPLE :Expand: (i) \({{\left( \frac{3}{5} \right)}^{4}}\)     (ii) \({{\left( \frac{-4}{7} \right)}^{5}}\)

SOLUTION

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

(ii) \({{\left( \frac{-4}{7} \right)}^{5}}=\frac{{{(-4)}^{5}}}{{{7}^{5}}}=\frac{(-4)\times (-4)\times (-4)\times (-4)}{7\times 7\times 7\times 7\times 7}\)

Numbers with exponent zero:

Can you tell what \(\frac{{{3}^{5}}}{{{3}^{5}}}\) is equal to?

\(\frac{{{3}^{5}}}{{{3}^{5}}}=\frac{3\times 3\times 3\times 3\times 3}{3\times 3\times 3\times 3\times 3}=1\)

by using laws of exponents

3 ⁵ ÷ 3⁵ = 3 ^{5 – 5} = 3⁰

So 3 ⁰ = 1

Can you tell what 7⁰ is equal to?

                   7 ³ ÷7 ³ = 7 ^{3 – 3} = 7

And                 \(\frac{{{7}^{3}}}{{{7}^{3}}}=\frac{7\times 7\times 7}{7\times 7\times 7}=1\)

Therefore         7 ⁰ = 1

Similarly           a ³ ÷ a ³ = a ^3–3 = a ⁰

And                  \({{a}^{3}}\div {{a}^{3}}=\frac{{{a}^{3}}}{{{a}^{3}}}=\frac{a\times a\times a}{a\times a\times a}=1\)

Thus                  a ⁰ = 1 (for any non-zero integer a)

So, we can say that any number (except 0) raised to the power (or exponent) 0 is 1

Exponents and Powers

MISCELLANEOUS EXAMPLES USING THE LAWS OF EXPONENTS

Let us solve some examples using rules of exponents developed.

EXAMPLE  Write exponential form for 8 × 8 × 8 × 8 taking base as 2. SOLUTION We have, 8 × 8 × 8 × 8 = 8⁴

But we know that 8 = 2 × 2 × 2 = 2³ Therefore 8 ⁴ = (2³ ) ⁴ = 2³ × 2³ × 2³ × 2³ = 2^{3 × 4} [You may also use (a ^m ) ⁿ = a ^mn] = 2 ¹²

EXAMPLE Simplify and write the answer in the exponential form

(i) \(\left( \frac{{{3}^{7}}}{{{3}^{2}}} \right)\times {{3}^{5}}\)  (ii) 2 ³ × 2 ² × 5 ⁵  (iii) (6² × 6⁴ ) ÷ 6³ (iv) [(2² ) ³ × 3⁶ ] × 5⁶ (v) 8 ² ÷ 2³

SOLUTION

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

= \(\left( {{3}^{7-2}} \right)\times {{3}^{5}}\)

= 3⁵×3⁵ = 3⁵⁺⁵ = 3¹⁰

(ii) 2 ³ × 2 ² × 5 ⁵ = 2³⁺² × 5 ⁵ = 2 ⁵ × 5 ⁵ = (2 × 5)⁵ = 10⁵ 

(iii)\(\begin{align}   & \left( {{6}^{2}}\times {{6}^{4}} \right)\div {{6}^{3}}={{6}^{2+4}}\div {{6}^{3}} \\  & =\frac{{{6}^{6}}}{{{6}^{3}}}={{6}^{6-3}}={{6}^{3}} \\ \end{align}\)

(iv) \(\begin{align}   & \left[ {{\left( {{2}^{2}} \right)}^{3}}\times {{3}^{6}} \right]\times {{5}^{6}}=\left[ {{2}^{6}}\times {{3}^{6}} \right]\times {{5}^{6}} \\  & ={{(2\times 3)}^{6}}\times {{5}^{6}} \\  & ={{(2\times 3\times 5)}^{6}}={{30}^{6}} \\ \end{align}\)

(v) 8 = 2 × 2 × 2 = 2³

Therefore 8 ² ÷ 2³ = (2³ ) ² ÷ 2³ = 2⁶ ÷ 2³ = 2^{6 –3} = 2³

EXAMPLE Simplify:

(i) \(\frac{{{12}^{4}}\times {{9}^{3}}\times 4}{{{6}^{3}}\times {{8}^{2}}\times 27}\) (ii) 2 ³ × a ³ × 5a ⁴ (iii) \(\frac{2\times {{3}^{4}}\times {{2}^{5}}}{9\times {{4}^{2}}}\)

SOLUTION (i) We have

\(\begin{align}   & \frac{{{12}^{4}}\times {{9}^{3}}\times 4}{{{6}^{3}}\times {{8}^{2}}\times 27}=\frac{{{({{2}^{2}}\times 3)}^{4}}\times {{({{3}^{2}})}^{3}}\times {{2}^{2}}}{{{(2\times 3)}^{3}}\times {{({{2}^{3}})}^{2}}\times {{3}^{3}}} \\  & =\frac{{{({{2}^{2}})}^{4}}\times {{(3)}^{4}}\times {{3}^{2\times 3}}\times {{2}^{2}}}{{{2}^{3}}\times {{3}^{3}}\times {{2}^{2\times 3}}\times {{3}^{3}}}=\frac{{{2}^{8}}\times {{2}^{2}}\times {{3}^{4}}\times {{3}^{6}}}{{{2}^{3}}\times {{2}^{6}}\times {{3}^{3}}\times {{3}^{3}}} \\  & =\frac{{{2}^{8+2}}\times {{3}^{4+6}}}{{{2}^{3+6}}\times {{3}^{3+3}}}=\frac{{{2}^{10}}\times {{3}^{10}}}{{{2}^{9}}\times {{3}^{6}}} \\ \end{align}\)

= 2^{10 – 9} × 3^{10 – 6}

= 2¹ × 3 ⁴ = 2 × 81 = 162

EXAMPLE Simplify:

(i) \(\frac{{{12}^{4}}\times {{9}^{3}}\times 4}{{{6}^{3}}\times {{8}^{2}}\times 27}\) (ii) 2 ³ × a ³ × 5a ⁴ (iii) \(\frac{2\times {{3}^{4}}\times {{2}^{5}}}{9\times {{4}^{2}}}\)

SOLUTION (i) We have

\(\begin{align}   & \frac{{{12}^{4}}\times {{9}^{3}}\times 4}{{{6}^{3}}\times {{8}^{2}}\times 27}=\frac{{{({{2}^{2}}\times 3)}^{4}}\times {{({{3}^{2}})}^{3}}\times {{2}^{2}}}{{{(2\times 3)}^{3}}\times {{({{2}^{3}})}^{2}}\times {{3}^{3}}} \\  & =\frac{{{({{2}^{2}})}^{4}}\times {{(3)}^{4}}\times {{3}^{2\times 3}}\times {{2}^{2}}}{{{2}^{3}}\times {{3}^{3}}\times {{2}^{2\times 3}}\times {{3}^{3}}}=\frac{{{2}^{8}}\times {{2}^{2}}\times {{3}^{4}}\times {{3}^{6}}}{{{2}^{3}}\times {{2}^{6}}\times {{3}^{3}}\times {{3}^{3}}} \\  & =\frac{{{2}^{8+2}}\times {{3}^{4+6}}}{{{2}^{3+6}}\times {{3}^{3+3}}}=\frac{{{2}^{10}}\times {{3}^{10}}}{{{2}^{9}}\times {{3}^{6}}} \\ \end{align}\)

= 2^{10 – 9} × 3^{10 – 6}

= 2¹ × 3 ⁴ = 2 × 81 = 162

(ii) 2 ³ × a ³ × 5a ⁴

= 2³ × a ³ × 5 × a ⁴

= 2 ³ × 5 × a ³ × a ⁴

= 8 × 5 × a ^{3 + 4} = 40 a ⁷

(iii) \(\begin{align}   & \frac{2\times {{3}^{4}}\times {{2}^{5}}}{9\times {{4}^{2}}}=\frac{2\times {{3}^{4}}\times {{2}^{5}}}{{{3}^{2}}\times {{\left( {{2}^{2}} \right)}^{2}}}=\frac{2\times {{2}^{5}}\times {{3}^{4}}}{{{3}^{2}}\times {{2}^{2\times 2}}} \\  & =\frac{{{2}^{1+5}}\times {{3}^{4}}}{{{2}^{4}}\times {{3}^{2}}}=\frac{{{2}^{6}}\times {{3}^{4}}}{{{2}^{4}}\times {{3}^{2}}}={{2}^{6-4}}\times {{3}^{4-2}} \\  & ={{2}^{2}}\times {{3}^{2}}=4\times 9=36 \\ \end{align}\)

Note: In most of the examples that we have taken in this Chapter, the base of a power was taken an integer. But all the results of the chapter apply equally well to a base which is a rational number.

Exponents and Powers

DECIMAL NUMBER SYSTEM

Let us look at the expansion of 47561, which we already know:

47561 = 4 × 10000 + 7 × 1000 + 5 × 100 + 6 × 10 + 1

We can express it using powers of 10 in the exponent form:

Therefore, 47561 = 4 × 10⁴ + 7 × 10³ + 5 × 10² + 6 × 10¹ + 1 × 10⁰

(Note 10,000 = 10⁴ , 1000 = 10³ , 100 = 10² , 10 = 10¹ and 1 = 10⁰ )

Let us expand another number:

104278 = 1 × 100,000 + 0 × 10,000 + 4 × 1000 + 2 × 100 + 7 × 10 + 8 × 1

= 1 × 10⁵ + 0 × 10⁴ + 4 × 10³ + 2 × 10² + 7 × 10¹ + 8 × 10⁰

= 1 × 10⁵ + 4 × 10³ + 2 × 10² + 7 × 10¹ + 8 × 10⁰

Notice how the exponents of 10 start from a maximum value of 5 and go on decreasing by 1 at a step from the left to the right upto 0.

Exponents and Powers

EXPRESSING LARGE NUMBERS IN THE STANDARD FORM

Let us now go back to the beginning of the chapter. We said that large numbers can be conveniently expressed using exponents. We have not as yet shown this. We shall do so now.

1. Sun is located 300,000,000,000,000,000,000 m from the centre of our Milky Way Galaxy.

2. Number of stars in our Galaxy is 100,000,000,000.

3. Mass of the Earth is 5,976,000,000,000,000,000,000,000 kg.

These numbers are not convenient to write and read. To make it convenient we use powers.

Observe the following:

59 = 5.9 × 10 = 5.9 × 10¹

590 = 5.9 × 100 = 5.9 × 10²

5900 = 5.9 × 1000 = 5.9 × 10³

59000 =5.9 × 10000 = 5.9 × 104 and so on.

       We have expressed all these numbers in the standard form. Any number can be expressed as a decimal number between 1.0 and 10.0 including 1.0 multiplied by a power of 10. Such a form of a number is called its standard form. Thus,

         5,985 = 5.985 × 1,000 = 5.985 × 103 is the standard form of 5,985.

       Note, 5,985 can also be expressed as 59.85 × 100 or 59.85 × 10² . But these are not the standard forms, of 5,985. Similarly, 5,985 = 0.5985 × 10,000 = 0.5985 × 10⁴ is also not the standard form of 5,985.

         We are now ready to express the large numbers we came across at the beginning of the chapter in this form.

The, distance of Sun from the centre of our Galaxy i.e., 300,000,000,000,000,000,000 m can be written as 3.0 × 100,000,000,000,000,000,000 = 3.0 × 1020 m

Now, can you express 40,000,000,000 in the similar way?

Count the number of zeros in it. It is 10.

So, 40,000,000,000 = 4.0 × 10¹⁰

        Mass of the Earth = 5,976,000,000,000,000,000,000,000 kg = 5.976 ×  10²⁴ kg

        Do you agree with the fact, that the number when written in the standard form is much easier to read, understand and compare than when the number is written with 25 digits? Now,

Mass of Uranus = 86,800,000,000,000,000,000,000,000 kg = 8.68 × 10²⁵ kg

      Simply by comparing the powers of 10 in the above two, you can tell that the mass of Uranus is greater than that of the Earth. The distance between Sun and Saturn is 1,433,500,000,000 m or 1.4335 × 10¹² m.

      The distance betwen Saturn and Uranus is 1,439,000,000,000 m or 1.439 × 10¹² m. The distance between Sun and Earth is 149, 600,000,000 m or 1.496 × 10¹¹m. Can you tell which of the three distances is smallest?

EXAMPLE Express the following numbers in the standard form:

(i) 5985.3 (ii) 65,950 (iii) 3,430,000 (iv) 70,040,000,000

SOLUTION

(i) 5985.3 = 5.9853 × 1000 = 5.9853 × 10³

(ii) 65,950 = 6.595 × 10,000 = 6.595 × 10⁴

(iii) 3,430,000 = 3.43 × 1,000,000 = 3.43 × 10⁶

(iv) 70,040,000,000 = 7.004 × 10,000,000,000 = 7.004 × 10¹⁰

       A point to remember is that one less than the digit count (number of digits) to the left of the decimal point in a given number is the exponent of 10 in the standard form. Thus, in 70,040,000,000 there is no decimal point shown; we assume it to be at the (right) end. From there, the count of the places (digits) to the left is 11. The exponent of 10 in the standard form is 11 – 1 = 10. In 5985.3 there are 4 digits to the left of the decimal point and hence the exponent of 10 in the standard form is 4 – 1 = 3.