## Percentage Increase

To increase a quantity by a percentage find the percentage of the quantity and add it to the original quantity.

Example 1: Increase 320 by 20%.

Solution. 20% of 320 = $$\frac{ 20}{ 100}$$ x 320 = 64

Therefore,            Increased amount = Rs 320 + Rs 64 = Rs 384.

## Percentage Decrease

To decrease a quantity by a percentage, find the percentage of the quantity and subtract it from the original quantity.

Example 2: Decrease 120 by $$12\frac{1}{2}$$ %.

Solution. $$12\frac{1}{2}$$% of 120

= Rs $$\frac{ 12.5}{ 100}$$ of 120

= Rs $$\frac{ 125 }{ 10 * 100}$$ x 120

= Rs 15

Therefore,       Decreased amount = Rs 120 — Rs15 = Rs 105.

## Percentage Change

 Percentage change = ( $$\frac{ Actual change (Increase or Decrease)}{ Original quantity}$$ x 100) % Percentage error = ($$\frac{ Error}{Actual Value}$$ x 100) %

Example 3: Sabita’s weight decreased from 80 kg to 40 kg. Find the percentage decrease.

Solution. Decrease in weight = (80 — 40) kg = 40 kg.

Therefore,        % decrease = $$\frac{ 40}{ 80}$$ x100% = 50%.

Example 4: The distance between two places was 200 km. It was measured as 300 km. Find the percentage error.

Solution. Error = 300 km — 200 km = 100 km.

Therefore,     %error = $$\frac{ error}{ actual value}$$ x 100%

= $$\frac{ 100}{ 200}$$ x 100 = 50%.

## Short Cut Method for Finding The Cubes Of a Two- Digit Number

We have :  $${ (a+b) }^{ 3 }$$ = $${ a }^{ 3 }$$ + 3 $${ a }^{ 2 }$$ b+ 3 a $${ b }^{ 2 }$$ + $${ b }^{ 3 }$$.
Method : For finding the cube of a two-digit number with the tens digit = a and the units digit b, we make four columns, headed by
$${ a }^{ 3 }$$, 3 $${ a }^{ 2 }$$ b, 3 a $${ b }^{ 2 }$$ and $${ b }^{ 3 }$$.
The rest of the procedure is the same as followed in squaring a number by the column method.

We simplify the working as :

Example 1: Find the value of $${ (29) }^{ 3 }$$ by the short-cut method.
Solution. Here,            a=2 and b=9

Therefore,        $${ (29) }^{ 3 }$$ = 24389.

Example 2: Find the value of $${ (71) }^{ 3 }$$ by the short-cut method.
Solution. Here, a = 7 and b = 1.

Therefore,     $${ (71) }^{ 3 }$$ = 357911.

## Procedure

Sometimes we are given two quantities and we want to find what per cent of one quantity is of the other quantity. In other words, we want to find how many hundredths of one quantity should be taken so that it is equal to the second quantity. In such type of problems, we proceed as discussed below:

Let a and b be two numbers and we want to know: what per cent of a is b ?

Let x% of a be equal to b. Then,

$$\frac{ x}{ 100}$$    x a = b

=> x = b x $$\frac{ 100}{ a }$$
=> x = $$\frac{ b}{ a}$$ x lOO

Thus, b is ($$\frac{ b}{ a }$$ x 100)% of a.

## Illustrative Examples:

Example 1: What per cent of 25 kg is 3.5 kg?

Solution. We have,

Required per cent = ( $$\frac{ 3.5 kg}{ 25 kg}$$ x 100) = $$\frac{ 3.5 * 100}{ 25}$$

= $$\frac{ 35 * 100}{ 250}$$

=$$\frac{ 35 * 2}{ 5}$$

= 7x 2

= 14

Hence, 3.5 kg is 14% of 25 kg.

Alternative Solution-

Let x% of 25 kg be 3.5 kg. Then,

x% of 25kg = 3.5kg

=> $$\frac{ x}{ 100}$$ x 25 = 3.5

=> x = $$\frac{ 3.5 * 100}{ 25}$$     [Multiplying both sides by $$\frac{ 100}{ 25}$$ ]

=> x = $$\frac{ 35 * 100}{ 250}$$ = $$\frac{ 35 * 2}{ 5}$$ = 7 x 2 = 14.

Example 2: Express 75 paise as a per cent of Rs 5.

Solution. We have, Rs 5 = 500 paise.

Let x% of Rs 5 be 75 paise. Then,

x% of Rs 5 = 75 paise

=> x% of 500 paise = 75 paise

=> $$\frac{ x}{ 100}$$ x 500 = 75

=> x = $$\frac{ 75 * 100}{ 500}$$

=> x = 15.

Hence, 15% of Rs 5 is 75 paise.

Alternative Solution-  The required per cent = ( $$\frac{ 75}{ 500}$$ x 100) % = 15%

Example 3 : Find 10% more than Rs 90.
Solution. We have,

10% of Rs 90 = Rs ( $$\frac{ 10 }{ 100}$$ x 90 ) = Rs 9

Therefore, 10% more than Rs 90 = Rs 90 + Rs 9 = Rs 99

## Steps involved in conversion of a ratio into per cent

STEP I– Obtain the ratio, say, a : b.

STEP II– Convert the given ratio into the fraction $$\frac{ a }{ b }$$

STEP III– Multiply the fraction obtained in step II by 100 and put per cent sign %.

Illustration 1: Express the following as per cents:
(i)  14 : 25                (ii) 5 : 6                (iii) 111 : 125

Solution. We have :

(i) 14 : 25 =$$\frac{ 14 }{ 25 }$$ = ($$\frac{ 14 }{ 25 }$$ x 100)% = 56%.

(ii) 5 : 6 = $$\frac{ 5 }{ 6 }$$ = ($$\frac{ 5 }{ 6 }$$ x 100)% = $$\frac{ 250 }{ 3 }$$% = $$83\frac{1}{3}$$%

(iii) 111 : 125 = $$\frac{ 111 }{ 125 }$$ = ($$\frac{ 111 }{ 125 }$$ x 100)% = 88.88%

## Steps involved in conversion of a per cent into ratio

STEP I– Obtain the per cent.

STEP II– Convert the given per cent into a fraction by dividing it by 100 and removing per cent sign %.

STEP III– Express the fraction obtained in step II in the simplest form.

STEP IV– Express the fraction obtained in step III as a ratio.

Illustration 2 : Express each of the following per cents as a ratio in the simplest form:

Solution.  We have:

(i) 36% = $$\frac{ 36 }{ 100 }$$ = 0.36

(ii) 5.4% = $$\frac{ 5.4 }{ 100 }$$ = $$\frac{ 54 }{ 1000 }$$ = 0.054

(iii) 0.25% = $$\frac{ 0.25 }{ 100 }$$ = $$\frac{ 25 }{ 10000 }$$ = 0.0025

(iv) 135% = $$\frac{ 135 }{ 100 }$$ = 1.35

## Steps involved in conversion of a per cent into a fraction

STEP I– Obtain the given per cent. Let it be x%.

STEP II– Drop the per cent sign (i.e %) and divide the number by 100. Thus, x% =  $$\frac{ x }{ 100 }$$

Illustration 1: Express the following per cents as fractions in the simplest forms:
(i) 57%            (ii) 36%             (iii) 115%
Solution. We have,

(i) 57% =  $$\frac{ 57 }{ 100 }$$

(ii) 36%=  $$\frac{ 36 }{ 100 }$$ =  $$\frac{ 9 }{ 25 }$$

(iii) 115% =  $$\frac{ 115 }{ 100 }$$ =  $$\frac{ 23 }{ 20 }$$

Illustration 2: Express each of the following per cents as fractions in the simplest
(i) 0.375%            (ii) 0.4%            (iii) 16%
Solution. We have,

(i) 0.375% = $$\frac{ 0.375 }{ 100 }$$ =  $$\frac{ 375 }{ 100000 }$$ =  $$\frac{ 3 }{ 800 }$$

(ii) 0.4% =  $$\frac{ 0.4 }{ 100 }$$ =  $$\frac{ 4 }{ 1000 }$$ =  $$\frac{ 1 }{ 250 }$$

(iii) $$16\frac{2}{3}$$ = $$\frac{ 50 }{ 3 }$$% = $$\frac { \frac { 50 }{ 3 } }{ 100 }$$ = $$\frac{ 50}{ 3 }$$ x $$\frac{ 1 }{ 100 }$$ = $$\frac{ 1 }{ 6 }$$

## Steps involved in conversion of a fraction into a percent

STEP I– Obtain the fraction. Let it be $$\frac{ a }{ b }$$

STEP II- Multiply the fraction by 100 and put the per cent sign% to obtain the required  percent. Thus, $$\frac{ 4 }{ 5 }$$ = ( $$\frac{ 4 }{ 5 }$$ x 100)%

Illustration 1: Express each of the following fractions as per cents:

(i) $$\frac{ 4 }{ 5 }$$    (ii) $$\frac{ 9 }{ 20 }$$    (iii) $$5\frac{1}{4}$$
Solution. We have,

(i) $$\frac{ 4 }{ 5 }$$ = ($$\frac{ 4 }{ 5 }$$  x 100)% = 80%

(ii) $$\frac{ 9 }{ 20 }$$ = ($$\frac{ 9 }{ 20 }$$  x 100)% = 45%

(iii) $$5\frac{1}{4}$$ = $$\frac{ 21 }{ 4 }$$ = ($$\frac{ 21 }{ 4 }$$ x 100)% = 525%

Illustration 2: Express each of the following into per cents:

(i) 0.375         (ii) 0.005         (iii) 2.45
Solution. We have,

(i)  0.375 =  $$\frac{ 375 }{ 1000 }$$% = ($$\frac{ 375 }{ 1000 }$$ x 100) = 37.5%

(ii)  0.005 = $$\frac{ 5 }{ 1000 }$$ = ($$\frac{ 5 }{ 1000 }$$ x 100)% = 0.5%

(iii)  2.45 = $$\frac{ 245 }{ 100 }$$ = ($$\frac{ 245 }{ 100 }$$ x 100)% = 245%

## Standard Form

A number written as ( m x $$10^n$$ ) is said to be in standard form if m is a decimal number such that 1 $$\le$$ m $$<$$10 and n is either a positive or a negative integer.

The standard form of a number is also known as Scientific notation.

## Expressing Very Large Numbers in Standard Form

In order to write large numbers in the standard form,following steps must be followed:

STEP I– Obtain the number and move the decimal point to the left till you get just one digit to the left of the decimal point.

STEP II– Write the given number as the product of the number so obtained and $$10^n$$ , where n is the number of places the decimal point has been moved to the left. If the given number is between 1 and 10, then write it as the product of the number itself and $$10^0$$ .

Illustrative Examples

Example 1: Express the following numbers in the standard form:

(i) 3,90,878    (ii) 3,186,500,000    (iii) 65,950,000

Solution.

(i) We have,

3,90,878 = 390878.00
Clearly, the decimal point is moved through five places to obtain a number in which there is just one digit to the left of the decimal point.

Therefore,    390878.00 = 3.90878 x $$10^5$$

(ii) We have,

3,186,500,000 = 3.186500000 x $$10^9$$
= 3.1865 x $$10^9$$
(iii) We have,

65,950,000 = 65,950,000.00

= 6.5950000 x $$10^7$$
= 6.595 x $$10^7$$

Example 2: The distance between sun and earth is (1.496 x $${10}^{11}$$) m and the distance between earth and moon is (3.84 x $$10^8$$) m. During solar eclipse moon comes in between earth and sun. At that time what Is the distance between moon and sun?

Solution.    Required distance

= {(1.496 x $${10}^{11}$$) – (3.84 x $$10^8$$) } m

= {$$\frac { 1496\times { 10 }^{ 11 } }{ { 10 }^{ 3 } }$$ – (3.84 x $$10^8$$)} m

= {1496 x $$10^8$$) – (3.84 x $$10^8$$)} m

= {(1496 – 3.84) x $$10^8$$)} m

= (1492.16 x $$10^8$$) m

Hence, the distance between moon and sun is (1492.16 x $$10^8$$ ) m.

Example 3: Write the following numbers in the usual form:

(i) 7.54 x $$10^6$$    (ii)2.514 x $$10^7$$

Solution.    We have

(i) 7.54 x $$10^6$$

= $$\frac{754}{100}$$ x $$10^6$$

= $$\frac{754 \times {10}^{6} }{{10}^{2}}$$

= 754 x $${10}^{(6-2)}$$

= (754 x $${10}^{4}$$ )

= (754 x 10000) = 7540000

(ii) 2.514 x $$10^7$$

= $$\frac{2514}{1000}$$ x $$10^7$$

= $$\frac{2514 \times {10}^{7} }{{10}^{3}}$$

= 2514 x $${10}^{(7-3)}$$

= (2514 x $${10}^{4}$$ )

= (2514 x 10000) = 25140000

## Expressing Very Small Numbers in Standard Form

In order to write very small numbers in the standard form,following steps must be followed:

STEP I- Obtain the number and count the number of decimal values after the decimal point. Consider it as n.

STEP II- Divide the number by $${10}^{n}$$). If the number is between 1 and 10, then write it as the product of the number itself and $${10}^{-n}$$

Example 1: Write the following numbers in the standard form:

(i) 0.000000059    (ii) 0.00000000526
Solution. We may write:

(i)    0.000000059

= $$\frac{59}{{10}^{9}}$$

=  $$\frac{5.9 \times 10}{{10}^{9}}$$

= $$\frac{5.9}{{10}^{8}}$$ = (5.9 x $${10}^{-8}$$)

(ii) 0.00000000526

= $$\frac{526}{{10}^{11}}$$

= $$\frac{5.26 \times 100}{{10}^{11}}$$

= $$\frac{5.26 \times {10}^{2}}{{10}^{11}}$$

= $$\frac{5.26}{{10}^{(11 – 2)}}$$

= $$\frac{5.26}{{10}^{9}}$$ = (5.26 x $${10}^{-9}$$)

Example 2: The size of a red blood cell is 0.000007 m and that of a plant cell Is 0.00001275 m. Show that a red blood cell is half of plant cell in size.

Solution.    We have,

Size of a red blood cell = 0.000007 m = $$\frac{7}{{10}^{6}}$$ m = (7 x $${10}^{-6}$$)

Size of a plant cell

= 0.00001275 m

= $$\frac{1275}{{10}^{8}}$$ m

= $$\frac{1.275 \times {10}^{3}}{{10}^{8}}$$ m

= $$\frac{1.275}{{10}^{(8-3)}}$$ m

= $$\frac{1.275}{{10}^{5}}$$ m = (1.275 x $${10}^{-5}$$) m

$$\frac{Size of a red blood cell}{Size of a plant cell}$$

= $$\frac{7 \times {10}^{-6}}{1.275 \times {10}^{-5}}$$

= $$\frac{7 \times {10}^{-6 + 5}}{1.275}$$

= $$\frac{7 \times {10}^{-1}}{1.275}$$

= $$\frac{7}{1.275 \times 10}$$

= $$\frac{7}{12.75}$$

= $$\frac{7}{13}$$ (nearly)

= $$\frac{1}{2}$$  (approximately)

Therefore,    size of a red blood cell = $$\frac{1}{2}$$ x (size of a plant cell)

Example 3: Express the following numbers in usual form:

(i) 3 x $${10}^{-3}$$    (ii) 2.34 x $${10}^{-4}$$
Solution.    We have,

(i) 3 x $${10}^{-3}$$

=  $$\frac{3}{{10}^{3}}$$

= $$\frac{3}{1000}$$ = 0.003

(ii) 2.34 x $${10}^{-4}$$

= $$\frac{234}{100}$$ x $$\frac{1}{{10}^{4}}$$

= $$\frac{234}{{10}^{2} \times {10}^{4}}$$

= $$\frac{234}{{10}^{6}}$$

= $$\frac{234}{1000000}$$ = 0.000234

## Formulas

Profit % = $$\frac{Profit}{ C.P.}$$ x 100

Loss % = $$\frac{ Loss}{C.P.}$$ x 100

S.P. = ($$\frac{100 + Profit \%}{ 100}$$) x C.P.

S.P. = ($$\frac{100 – Loss \%}{ 100}$$) x C.P.

## Illustrative Examples

Example 1: Kirpal bought a certain number of apples at Rs 75 per score and sold them at a profit of 40%. Find the selling price per apple.

Solution. C.P. of one score i.e.,  20 apples = Rs 75, profit = 40%
S.P. of 20 apples= ( 1 + $$\frac{40}{100}$$ )of Rs 75

= Rs ( $$\frac{140}{100}$$ x 75 ) = Rs 105

S.P. of one apple = Rs $$\frac{105}{20}$$

= Rs $$\frac{21}{4}$$ = Rs 5.25

Example 2: Bashir bought an article for Rs 1215 and spent Rs 35 on its transportation. At what price should he sell the article to have a gain of 16%?

Solution. The effective cost price of the article is equal to the price at which it was bought plus the transportation charge.

C.P. of the given article = Rs (1215 + 35) = Rs 1250
Gain percent = 16%

Gain = 16% of cost price = Rs ($$\frac{16}{100}$$ x 1250) = Rs 200

S.P. = C.P. + Gain = Rs 1250 + Rs 200 = Rs 1450

Example 3: Krishnamurti bought oranges at Rs 5 a dozen. He had to sell them at a loss of 4%. Find the selling price of one orange.

Solution. We have, C.P. of one dozen oranges = Rs 5.

Loss percent = 4%

Loss = 4 % of Rs 5 = Re($$\frac{4}{100}$$ x 5) = Re ($$\frac{1}{5}$$)

S.P. = C.P. — Loss = Rs (5 – $$\frac{1}{5}$$) = Rs $$\frac{24}{5}$$

Thus, S.P. of one dozen oranges = Rs $$\frac{24}{5}$$

Therefore, S.P.of an orange = Re ($$\frac{24}{5}$$ x $$\frac{1}{12}$$)
= Re $$\frac{2}{5}$$
= $$\frac{2}{5}$$ x 100 paise
= 40 paise

## Definitions

Compound Interest: If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half year or a quarter of a year etc) so that the amount ( = Principal + Interest) at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals, calculated in this way is called the compound interest and is abbreviated as C.I.

Clearly, compound interest at the end of certain specified period is equal to the difference between the amount at the end of the period and the original principal i.e.
C.I. = Amount — Principal

Conversion Period: The fixed interval of time at the end of which the interest is calculated and added to the principal at the beginning of the interval is called the conversion period.

In other words, the period at the end of which the interest is compounded is called the conversion period.

When the interest is calculated and added to the principal every six months, the conversion period is six months. Similarly, the conversion period is 3 months when the interest is calculated and added quarterly.

## Finding CI When Interest Is Compounded Annually

when Interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.

Example 1: Maria invests Rs 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate:
(i) The amount standing to her credit at the end of second year.
(ii) The interest for the third year.
Solution.    (i) We have,
Principal for the first year Rs 93750

Rate of interest = 9.6% per annum.

Therefore,          Interest for the first year = Rs ($$\frac{93750 \times 9.6 \times 1}{100}$$) = Rs 9000

Amount at the end of the first year = Rs 93750 +Rs 9000

= Rs 102750

Principal for the second year = Rs 102750

Interest for the second year = Rs ($$\frac{102750 \times 9.6 \times 1}{100}$$) = Rs 9864

Amount at the end of second year = Rs 102750 + Rs 9864

= Rs 112614

(ii) Principal for the third year = Rs 112614

Interest for the third year =Rs ($$\frac{112614 \times 9.6 \times 1}{100}$$ ) = Rs 10810.94

Example 2: Find the compound interest on Rs 25000 for 3 years at 10% per annum, compounded annually.

Solution.    Principal for the first year = Rs 25000

Interest for the first year  = ($$\frac{25000 \times 10 \times 1}{100}$$) = Rs 2500

Amount at the end of the first year = (25000 + 2500) = Rs 27500

Principal for the second year = Rs 27500

Interest for the second year = ($$\frac{27500 \times 10 \times 1}{100}$$) = Rs 2750

Amount at the end of the second year = (27500 + 2750) = Rs 30250.

Principal for the third year = Rs 30250.

Interest for the third year = ($$\frac{30250 \times 10 \times 1}{100}$$) = Rs 3025

Amount at the end of the third year = (30250 + 3025) = Rs 33275.

Therefore, compound interest = (33275 — 25000) = Rs 8275

## Finding CI When Interest Is Compounded Half-Yearly

If the rate of Interest is R% per annum then it is clearly ($$\frac{R}{2}$$)% per half-year.
The amount after the first half-year becomes the principal for the next half-year, and so on.

Example 3: Find the compound interest on Rs 5000 for 1 year at 8% per annum, compound half-yearly.

Solution.    Rate of interest             = 8% per annum = 4% per half-year.

Time                                          = 1 year = 2 half-years.

Original principal                    = Rs 5000.

Interest for the first half-year = ($$\frac{5000 \times 4 \times 1}{100}$$) = Rs 200.

Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = Rs 5200.

Interest for the second half-year = (($$\frac{5200 \times 4 \times 1}{100}$$)) = Rs 208.

Amount at the end of the second half-year = Rs (5200 + 208) = Rs 5408.
Therefore,        compound interest = Rs (5408 — 5000) = Rs 408.

Example 4: Find the compound interest on Rs 8000 for $$1\frac{1}{2}$$ years at 10% per annum, interest being payable half-yearly.

Solution.    We have,

Rate of interest                = 10% per annum = 5% per half-year.

Time                                  = $$1\frac{1}{2}$$ years = $$\frac{3}{2}$$ x 2 = 3 half- years

Original principal             = Rs 8000

Interest for the first half-year = Rs ( $$\frac{8000 \times 5 \times 1}{100}$$) = Rs 400

Amount at the end of the first half-year = Rs 8000 + R 400 = Rs 8400
Principal for the second half-year = Rs 8400

Interest for the second half-year = Rs ($$\frac{8400 \times 5 \times 1}{100}$$) = Rs 420

Amount at the end of the second half-year = Rs 8400 + Rs 420 = Rs 8820
Principal for the third half-year = Rs 8820

Interest for the third half-year = Rs ($$\frac{8820 \times 5 \times 1}{100}$$) = Rs 441

Amount at the end of third half-year = Rs 8820 + Rs 441 = Rs 9261
Therefore,        Compound interest = Rs 9261 — Rs 8000 = Rs 1261

## Finding CI When Interest Is Compounded Quarterly

If the rate of interest is R % per annum and the interest is compounded quarterly, then it is $$\frac{R}{4}$$ %  per quarter.

Example 5: Find the compound interest on Rs 10000 for 1 year at 20% per annum compounded quarterly.

Solution.    We have,

Rate of interest             = 20% per annum = $$\frac{1}{5}$$% = 5% per quarter

Time                         = 1 year = 4 quarters.

Principal for the first quarter = Rs 10000

Interest for the first quarter = Rs ($$\frac{10000 \times 5 \times 1}{100}$$) = Rs 500

Amount at the end of first quarter = Rs 10000 + Rs 500 = Rs 10500

Principal for the second quarter = Rs 10500

Interest for the second quarter = Rs ($$\frac{10500 \times 5 \times 1}{100}$$) = Rs 525

Amount at the end of second quarter = Rs 10500 + Rs 525 = Rs 11025
Principal for the third quarter = Rs 11025

Interest for the third quarter = Rs ($$\frac{11025 \times 5 \times 1}{100}$$) = Rs 551.25

Amount at the end of the third quarter = Rs 11025 + Rs 551.25

= Rs 11576.25

Principal for the fourth quarter = Rs 11576.25

Interest for the fourth quarter = Rs ($$\frac{11576.25 \times 5 \times 1}{100}$$) = Rs 578.8125

Amount at the end of fourth quarter = Rs 11576.25 + Rs 578.8125

= Rs 12155.0625

Therefore,        Compound interest = Rs 12155.0625 — Rs 10000

= Rs 2155.0625

= Rs 2155.06

## Formulas

Profit = S.P. –C.P.

Loss = C.P. – S.P.

Profit % = ( $$\frac{Profit}{C.P.}$$ x 100) %

Loss % = ( $$\frac{Loss}{C.P.}$$ x 100) %

## Illustrative Examples

Example 1: John bought a watch for Rs 540 and sold it for Rs 585. Find his profit and profit percentage.

Solution. C.P. of the watch = Rs 540, S.P. of the watch = Rs 585

Profit = S.P. — C.P.
= Rs 585 — Rs 540 = Rs 45

Profit percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %

= ($$\frac{45}{540}$$ X 100 ) %

=  $$\frac{100}{12}$$ %

= $$\frac{25}{3}$$ %
= $$8\frac{1}{3}$$ %

Example 2: By selling a bike for Rs 22464, Ansari incurs a loss of Rs 1536. Find his loss percenta

Solution. S.P. of the bike = Rs 22464, loss = Rs 1536
C.P. of the bike = S.P. + loss = Rs 22464 + Rs 1536 = Rs 24000

Loss percentage = ( $$\frac{Loss}{C.P.}$$ x 100) %

= ($$\frac{1536}{24000}$$ X 100 ) %

=  $$\frac{1536}{240}$$ %
= $$\frac{64}{10}$$ %
= 6.4 %

Example 3: B ijoy bought bananas at the rate of 5 for Rs 4 and sold them at the rate of 4 for Rs 5. Calculate his gain percentage.

Solution.

C.P. of 5 bananas = Rs 4

C.P. of 1 banana = Rs $$\frac{4}{5}$$ = Rs 0.80

S.P. of 4 bananas = Rs 5

S.P. of 1 banana = Rs $$\frac{5}{4}$$ = Rs 1.25

Therefore,        Gain on the sale of one banana = S.P. — C.P. = Rs 1.25 — Rs 0.80 = Rs 0.45

Gain percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %

= ($$\frac{0.45}{0.80}$$ X 100 ) %

=  $$\frac{145}{80}$$ %
= $$\frac{225}{4}$$ %
= 56.25 %

## First Law

If a is any non-zero rational number and m, n are natural numbers, then

$${ a }^{ m }$$ x $${ a }^{ n }$$ = $${ a }^{ m + n }$$
Generalised form of above law:

If a is a non-zero rational number and m, n, p are natural numbers, then,

$${ a }^{ m }$$ x $${ a }^{ n }$$ x $${ a }^{ p }$$ = $${ a }^{ m + n + p }$$

Illustration : Simplify and write the answer of each of the following in exponential form:

(i) $${ 4 }^{ 2 }$$ x $${ 4 }^{ 3 }$$        (ii) $${ 2 }^{ 2 }$$ x $${ 2 }^{ 3 }$$ x $${ 2 }^{ 4 }$$
(iii) $${ 6 }^{ x }$$ x $${ 6 }^{ 3 }$$     (iv) $${ (\frac { 3 }{ 2 } ) }^{ 3 }$$ x $${ (\frac { 3 }{ 2 } ) }^{ 6 }$$

Solution. Using first law of exponents, We have

(i) $${ 4 }^{ 2 }$$ x $${ 4 }^{ 3 }$$ = $${ 4 }^{ 2 + 3 }$$ = $${ 4 }^{ 5 }$$

(ii) $${ 2 }^{ 2 }$$ x $${ 2 }^{ 3 }$$ x $${ 2 }^{ 4 }$$ = $${ 2 }^{ 2 + 3 + 4 }$$ = $${ 2 }^{ 9 }$$

(iii) $${ 6 }^{ x }$$ x $${ 6 }^{ 3 }$$ = $${ 6 }^{ x + 3 }$$

(iv) $${ (\frac { 3 }{ 2 } ) }^{ 3 }$$ x $${ (\frac { 3 }{ 2 } ) }^{ 6 }$$= $${ (\frac { 3 }{ 2 } ) }^{ 6 + 3 }$$ = $${ (\frac { 3 }{ 2 } ) }^{ 9 }$$

## Second Law

If a is any non-zero rational number and m and n are natural numbers such that m > n, then

$${ a }^{ m }$$ $$\div$$ $${ a }^{ n }$$ = $${ a }^{ m – n }$$ or $$\frac { { a }^{ m } }{ { a }^{ n } }$$ = $${ a }^{ m – n }$$

Illustration : Simplify and write each of the following in exponential form:

(i) $${ 8 }^{ 6 }$$ $$\div$$ $${ 8 }^{ 3 }$$        (ii) $${ (-5) }^{ 10 }$$ ÷ $${ (-5) }^{ 4 }$$
(iii) $${ (\frac { -3 }{ 5 } ) }^{ 6 }$$ ÷ $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$
Solution. Using second law of exponents, We have

(i) $${ 8 }^{ 6 }$$ $$\div$$ $${ 8 }^{ 3 }$$

= $$\frac { { 8 }^{ 6 } }{ { 8 }^{ 3 } }$$
=  $${ 8 }^{ 6 – 3 }$$
=  $${ 8 }^{ 3 }$$
(ii) $${ (-5) }^{ 10 }$$ ÷ $${ (-5 )}^{ 4 }$$

= $$\frac { { (-5) }^{ 10 } }{ {(-5) }^{ 4 } }$$

= $${ (-5) }^{ 10 – 4 }$$
= $${ (-5) }^{ 6 }$$
(iii) $${ (\frac { -3 }{ 5 } ) }^{ 6 }$$ ÷ $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$

= $${ \frac { { (\frac { -3 }{ 5 } })^{ 6 } }{ ({ \frac { -3 }{ 5 } ) }^{ 3 } } }$$
= $${ (\frac { -3 }{ 5 } ) }^{ 6 – 3 }$$
= $${ (\frac { -3 }{ 5 } ) }^{ 3 }$$

## Third Law

If a is any rational number different from zero and m, n are natural numbers, then

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

Illustration: Simplify and write each of the following in exponential form:

(i) $${ { (3 }^{ 2 }) }^{ 4 }\quad$$    (ii) $${ { ((-2) }^{ 4 }) }^{ 2 }\quad$$
(iii) $${ { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 }$$    (iv) $${ ({ 4 }^{ 2 }) }^{ 3 }\quad$$ x $$({ { 4}^{ 4 }) }^{ 2 }\quad$$
Solution. Using third law of exponents, We have

(i) $${ { (3 }^{ 2 }) }^{ 4 }\quad$$

= $${ 3 }^{ 2 \times 4 }$$
= $${ 3 }^{ 8 }$$
(ii) $${ { ((-2) }^{ 4 }) }^{ 2 }\quad$$

= $${ (-2) }^{ 4 \times 2 }$$

= $${ (-2) }^{ 8 }$$
(iii) $${ { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\} }^{ 4 }$$

= $${ (\frac { 3 }{ 5 } ) }^{ 3 \times 4 }$$
= $${ (\frac { 3 }{ 5 } ) }^{ 12 }$$

(iv) $${ ({ 4 }^{ 2 }) }^{ 3 }\quad$$x $${ ({ 4}^{ 4 }) }^{ 2 }\quad$$

= $$({ 4 }^{ 2 \times 3})$$ x $$({ 4 }^{ 4 \times 2 })$$
= $${ 4 }^{ 6 }$$ x $${ 4 }^{ 8 }$$
= $${ 4 }^{ 6 + 8 }$$
= $${ 4 }^{ 14 }$$

## Fourth Law

If a, b are non-zero rational numbers and n is a natural number, then

$${ a }^{ n }$$ x $${ b }^{ n }$$ = $${ (ab) }^{ n }$$
Generalised form of above law:

If a, b, c are non-zero rational numbers and n is a natural number, then

$${ a }^{ n }$$ x $${ b }^{ n }$$ x $${ c }^{ n }$$ = $${(abc)}^{ n }$$

Illustration: Express each of the following products of powers as the exponent of a rational number:

(i) $${ 2 }^{ 4 }$$ x $${ 5 }^{ 4 }$$    (ii) $${(-3) }^{ 3 }$$ x $${ (-2) }^{ 3 }$$
(iii) $${ 3 }^{ 2 }$$ x $${ x }^{ 2 }$$ x $${ y }^{ 2 }$$    (iv) $${ (\frac { 3 }{ 2 } ) }^{ 2 }$$ x $${ (\frac { 2 }{ 5 } ) }^{ 2 }$$
Solution. We have,

(i) $${ 2 }^{ 4 }$$ x $${ 5 }^{ 4 }$$

= $${ (2 \times 5 )}^{ 4 }$$
= $${ 10 }^{ 4 }$$
(ii) $${(-3) }^{ 3 }$$ x $${ (-2) }^{ 3 }$$

= $${ ((-3) \times (-2) )}^{ 3 }$$

= $${ 6 }^{ 3 }$$

(iii) $${ 3 }^{ 2 }$$ x $${ x }^{ 2 }$$ x $${ y }^{ 2 }$$

= $${( 3 \times x \times y )}^{ 2 }$$
= $${ (3xy) }^{ 2 }$$

(iv) $${ (\frac { 3 }{ 2 } ) }^{ 2 }$$ x $${ (\frac { 2 }{ 5 } ) }^{ 2 }$$

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

## Fifth Law

If a and b are non-zero rational numbers and n is a natural number, then

$$\frac { { a }^{ n } }{ { b }^{ n } }$$ = $${ (\frac { a }{ b } ) }^{ n }$$

Illustration: Write each of the following in the form $$\frac{p}{q}$$

(i) $${ (\frac { 2 }{ 5 } ) }^{ 3 }$$    (ii) $${ (\frac { -3 }{ 2 } ) }^{ 4 }$$
Solution. We have,

(i) $${ (\frac { 2 }{ 5 } ) }^{ 3 }$$

= $$\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$\frac{2 \times 2 \times 2}{5 \times 5 \times 5 }$$
= $$\frac{8}{125}$$
(ii) $${ (\frac { -3 }{ 2 } ) }^{ 4 }$$

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

= $$\frac{(-3) \times (-3) \times (-3) \times (-3)}{2 \times 2 \times 2 \times 2}$$
= $$\frac{81}{16}$$

## Illustrative Examples

Example 1: Use the laws of exponents to simplify the following :

(i) $${ [{ (2^{ 3 }) }^{ 4 }] }^{ 5 }$$    (ii) $$[{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 }$$
(iii) $${ 81 }^{ -1 }$$ x $${ 3 }^{ 5 }$$    (iv) $${ (\frac { 2 }{ 3 } ) }^{ 0 }$$ + $${ (\frac { 2 }{ 3 } ) }^{ -2 }$$
Solution. We have,

(i) $${ [{ (2^{ 3 }) }^{ 4 }] }^{ 5 }$$

= $${ [{ 2^{ 3 \times 4 }}] }^{ 5 }$$
= $${ [{ 2^{ 12 }}] }^{ 5 }$$
= $${ [{ 2^{ 12 \times 5}}] }$$
= $${ 2 }^{ 60 }$$
(ii) $$[{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 }$$

= $${ [ 3^{ 6 – 4 }]^3}$$

= $${ [{ 3^{ 2 }}] }^{ 3 }$$
= $${ 3 }^{ 2 \times 3 }$$
= $${ 3 }^{ 6 }$$

(iii) $${ 81 }^{ -1 }$$ x $${ 3 }^{ 5 }$$

= $${ { (3 }^{ 4 }) }^{ -1 }\quad$$ x $${ 3 }^{ 5 }$$
= $${ 3 }^{ -4 + 5 }$$
= $${ 3 }^{ 1 }$$
= 3

(iv) $${ (\frac { 2 }{ 3 } ) }^{ 0 }$$ + $${ (\frac { 2 }{ 3 } ) }^{ -2 }$$

= 1 + $${ \frac { 1 }{ ({ \frac { 2 }{ 3 } ) }^{ 2 } } }$$
= $${ \frac { 1 }{ \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } } } }$$
= $${ \frac { 1 }{ \frac { 4 }{ 9 } } }$$
= 1 + $$\frac{9}{4}$$
= $$\frac{13}{4}$$

Example 2: Simplify and write the answer in the exponential form:

(i) $${ ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad }$$    (ii) $${ (-4) }^{ 3 }$$ x $${ (5) }^{ -3 }$$ x $${ (-5) }^{ -3 }$$

Solution.

(i) We have,

$${ ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad }$$
= $${ (\frac { { 2 }^{ 5 } }{ { 2 }^{ 8 } } ) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ { (2 }^{ 5 – 8 }) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ { (2 }^{ -3 }) }^{ 5 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -3 \times 5 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -15 }$$ x $${ (2) }^{ -5 }$$
= $${ (2) }^{ -15-5 }$$ = $${ (2) }^{ -20 }$$

(ii) We have,

$${ (-4) }^{ 3 }$$ x $${ (5) }^{ -3 }$$ x $${ (-5) }^{ -3 }$$
= $${ [-4 \times 5 \times (-5)] }^{ -3 }$$
= $${ (100) }^{ -3 }$$
= $${ { (10 }^{ 2 }) }^{ -3 }\quad$$
= $${ (10) }^{ 2 \times -3 }$$ = $${ (10) }^{ -6 }$$

Example 3: Simplify

(i) $${ ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }}$$    (ii) $${ ({ { 5 }^{ 3 } \times { 3 }^{ -1 }) }^{ -1 }}$$ ÷ $${ 6 }^{ -1 }$$

Solution.

(i) $${ ({ { 4 }^{ -1 } \div { 3 }^{ -1 }) }^{ -2 }}$$

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

=  $${ (\frac { 3 }{ 4 })^{-2}}$$

= $${ (\frac { 4 }{ 3 })^{2}}$$

= $$\frac{16}{9}$$

(ii) $${ ({ { 5 }^{ -1 } \times { 3 }^{ -1 }) }^{ -1 }}$$ ÷ $${ 6 }^{ -1 }$$

= $${ (\frac { 1 }{ 5 } \times \frac { 1 }{ 3 } ) }^{ -1 }$$ ÷ $$\frac{1}{6}$$

= $${ (\frac { 1 }{ 15 })^ {-1} }$$ ÷ $$\frac{1}{6}$$

= $$\frac{15}{1}$$ ÷ $$\frac{1}{6}$$

= $$\frac{15}{1} \times \frac{6}{1}$$

= 90

Example 4: Find the value of m if $${ (\frac { 2 }{ 9 })^ {3} }$$ x $${ (\frac { 2 }{ 9 })^ {-6} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

Solution.

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

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

=>    $${ (\frac { 2 }{ 9 })^ {-3} }$$ = $${ (\frac { 2 }{ 9 })^ {2m-1} }$$

In an equation, when bases on both sidess are equal, their powers must also be equal.

Therefore, 2m – 1 = -3 or 2m = -3 + 1

=>    2m = -2

=>    m = $$\frac{-2}{2}$$ = –1

Example 5: What number should $${ (\frac { 2 }{ 3 })^ {-2} }$$ be multiplied so that the product is $${ (\frac { 4 }{ 27 })^ {-1} }$$ ?

Solution.

$$x$$ x $${ (\frac { 2 }{ 3 })^ {-2} }$$  = $${ (\frac { 4 }{ 27 })^ {-1} }$$

=>    $$x$$ x  $${ (\frac { 3 }{ 2 })^ {2} }$$ = $$\frac{27}{4}$$

=>    $$x$$ x $$\frac{9}{4}$$ = $$\frac{27}{4}$$

=>    $$x$$ = $$\frac{27}{4}$$ x $$\frac{4}{9}$$

=>    $$x$$ = 3.

Hence, the required number is 3.

Example 6: Simplify

$$\frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 }$$
Solution.    We have,

$$\frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 }$$
= $$\frac { ({ 2 }^{ 2 }\times 3)^{4} \times ({ 3 }^{ 2 }) ^ {3} \times {2}^{2} }{ { {2}^ {3}\times 3 }^{ 3 }\times ({ 2 }^{ 3 })^{2}\times {3}^{3}) }$$

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

= $$\frac { { 2 }^{ 8 }\times { 3 }^{ 4 }\times {3}^{6} \times {2}^{2} }{ { 2 }^{ 3 }\times { 3 }^{ 3 }\times {2}^{6} \times {3}^{3} }$$

= $$\frac { ({ 2 }^{ 8 }\times { 2 }^{ 2 } )\times ({3}^{6} \times {3}^{4}) }{ ({ 2 }^{ 3 }\times { 2 }^{ 6 }\times ({3}^{3} \times {3}^{3}) }$$

= $$\frac { { 2 }^{ 8+2 }\times { 3 }^{ 4+6 } }{ { 2 }^{ 3+6 }\times { 3 }^{ 3+3 } }$$

= $$\frac { { 2 }^{ 10 }\times { 3 }^{ 10 } }{ { 2 }^{ 9 }\times { 3 }^{ 6 } }$$

= $$\frac { { 2 }^{ 10 } }{ { 2 }^{ 9 } }$$ x $$\frac { { 3 }^{ 10 } }{ { 3 }^{ 6 } }$$

= $${ 2 }^{ 10-9 }$$ x $${ 3 }^{ 10-6 }$$

= $${ 2 }^{ 1 }$$ x $${ 3 }^{ 4 }$$

= 2 x 81 = 162

Example 7: Find the values of n in each of the following:

(i) $${ ({ 2 }^{ 2 }) }^{ n }$$ = $${ ({ 2 }^{ 3 }) }^{ 4 }$$    (ii) $${ 2 }^{ 5n }$$ ÷ $${ 2 }^{ n }$$ = $${ 2 }^{ 4 }$$

Solution.

(i) We have,

$${ ({ 2 }^{ 2 }) }^{ n }$$ = $${ ({ 2 }^{ 3 }) }^{ 4 }$$
=>    $${ 2 }^{ 2n }$$ = $${ 2 }^{ 3 \times 4 }$$

=>    $${ 2 }^{ 2n }$$ = $${ 2 }^{ 12 }$$

=>    2n = 12        [On equating the exponents]

=>    n = $$\frac{12}{2}$$ = 6

(ii) We have,

$${ 2 }^{ 5n }$$ ÷ $${ 2 }^{ n }$$ = $${ 2 }^{ 4 }$$
=>    $$\frac { { 2 }^{ 5n } }{ { 2 }^{ n } }$$ = $${ 2 }^{ 4 }$$

=>    $${ 2 }^{ 5n-n }$$ = $${ 2 }^{ 4 }$$    [since, $$\frac { { a }^{ m } }{ { a }^{ n } }$$ = $${ a }^{ m-n }$$ ]

=>    $${ 2 }^{ 4n }$$ = $${ 2 }^{ 4 }$$

=>     4n = 4         [On equating the exponents]

=>    n = $$\frac{4}{4}$$ = 1

Example 8: If $${25}^{n-1}$$ + 100 = $${5}^{2n-1}$$, find the value of n.

Solution.    We have,

$${25}^{n-1}$$ + 100 = $${5}^{2n-1}$$

=>    $${5}^{2n-1}$$ – $${5}^{2n-1}$$ = 100

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { { 25 }^{ n } }{ { 25} }$$ = $${10}^{2}$$

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { ({ 5 } ^{2})^{ n } }{ { 25} }$$ = $$({2 \times 5}) ^{2}$$

=>    $$\frac { { 5 }^{ 2n } }{ { 5} }$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$\frac{1}{5}$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{1}{5})$$ – $$\frac { { 5 }^{ 2n } }{ { 25} }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{5 – 1}{25})$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$(\frac{4}{25})$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $$\frac { { 2 }^{ 2 } }{ { 5 }^{ 2 } }$$ = $${2}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ x $${2}^{2}$$ = $${2}^{2} \times {5}^{2} \times {5}^{2}$$

=>    $${5}^{2n}$$ = $$\frac { { 2 }^{ 2 } \times {5}^{2} \times {5}^{2} }{ { 2 }^{ 2 } }$$

=>    $${5}^{2n}$$ = $${2}^{2 – 2}$$ x $${5}^{2 + 2}$$

=>    $${5}^{2n}$$ = $${2}^{0}$$ x $${5}^{4}$$

=>    $${5}^{2n}$$ = $${5}^{4}$$

=>    2n = 4     [On equating the exponents]

=>    n = $$\frac{4}{2}$$ = 2

## Definition

If a is any real number and n is a natural number, then $${ a }^{ n }$$ = a x a x a…. n times

where a is called the base, n is called the exponent or index and  $${ a }^{ n }$$ is the exponential expression. $${ a }^{ n }$$ is read as ‘a raised to the power n’ or ‘a to the power n’ or simply ‘a power n’.

For zero power, we have :

$${ a }^{ 0 }$$ = 1 (where a $$\neq$$ 0)

For example :

(i) $${ 7 }^{ 0 }$$ =  1    (ii) $${ (-\frac { 2 }{ 3 } ) }^{ 0 }$$ = 1    (iii) $$( { \sqrt { 7 } })^{ 0 }$$ = 1

For negative powers, we have :

$$\sqrt [ n ]{ { a }}$$ = $$\frac { 1 }{ { a }^{ n } }$$ and $$\frac { 1 }{ { a }^{ -n } }$$ = $${ a }^{ n }$$

For example:
(i) $${ 5 }^{ -2 }$$ = $$\frac { 1 }{ { 5 }^{ 2 } }$$
(ii) $${ -2 }^{ -3 }$$ = $$\frac { 1 }{ { -2 }^{ 3 } }$$
(iii) $$\frac { 1 }{ { 2 }^{ -5 } }$$ = $${ 2 }^{ 5 }$$

For fractional indices,  we have :

$${ \sqrt { a} }^{ n }$$ = $${ a }^{ \frac { 1 }{ n } }$$ and $$\sqrt [ n ]{ { a }^{ m } }$$ = $${ a }^{ \frac { m }{ n } }$$

For example:

(i) $${ \sqrt { 3 } }$$ = $${ 3 }^{ \frac { 1 }{ 2 } }$$
(ii) $${ \sqrt { 8 } }^{ 3 }$$ = $${ 8 }^{ \frac { 1 }{ 3} }$$
(iii) $$\sqrt [ 4 ]{ { 5 }^{ 3 } }$$ = $${ 5 }^{ \frac { 3 }{ 4 } }$$

## Finding the value of the Number given in the Exponential Form

Example 1: Find the value of each of the following:

(i) $${ 12 }^{ 2 }$$    (ii) $${ 8 }^{ 3 }$$    (iii) $${ 4 }^{ 4 }$$

Solution.

(i) We have,

$${ 12 }^{ 2 }$$ = 12 x 12 = 144
(ii) We have,

$${ 8 }^{ 3 }$$ = 8 x 8 x 8

= (8 x 8) x 8
= 64 x 8
= 512
(iii) We Have,

$${ 4 }^{ 4 }$$= 4 x 4 x 4 x 4

= (4 x 4 ) x 4 x 4
= (16 x 4) x 4
= 64 x 4
= 256

Example 2: Simplify:

(i) 2 x $${ 10 }^{ 3 }$$    (ii) $${ 5 }^{ 2 }$$ x $${ 4 }^{ 2 }$$    (iii) $${ 3 }^{ 3 }$$ x 4

Solution.

(i) We have,

2 x $${ 10 }^{ 3 }$$ = 2 x 1000 = 2000     [since $${ 10 }^{ 3 }$$=10 x10 x 10 = 1000]

(ii) We have,

$${ 5 }^{ 2 }$$ x $${ 4 }^{ 2 }$$

= 25 x 16 = 400

(iii) We have,

$${ 3 }^{ 3 }$$ x 4  = 27 x 4 = 108

## Expressing Numbers in Exponential Form

Example 1: Express each of the following in exponential form:

(i) (-4) x (-4) x (-4) x (-4) x (-4)    (ii) $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$
Solution. We have,

(i)  (-4) x (-4) x (-4) x (-4) x (-4) = $${ -4}^{ 5 }$$

(ii) $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ = $$({ \frac { 2 }{ 5 } ) }^{ 4 }$$

Example 2: Express each of the following in exponential form:

(i) 3 x 3 x 3 x a x a    (ii) a x a x a x a x a x a x b x b x b c x c x c x c
(iii) b x b x b x $$\frac{2}{5}$$ x $$\frac{2}{5}$$
Solution. We have,

(i) 3 x 3 x 3 x a x a = $${ 3 }^{ 3 }$$ x $${ a }^{ 2 }$$

(ii) a x a x a x a x a x a x b x b x b x c x c x c x c = $${ a }^{ 6 }$$ x $${ b }^{ 3 }$$ x $${ c }^{ 4 }$$

(iii) b x b x b x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ = $${ a }^{ 3 }$$ x $$({ \frac { 2 }{ 5 } ) }^{ 2 }$$

Example 3: Express each of the following numbers in exponential form:

(i) 128    (ii) 243    (iii) 3125
Solution.

(i) We have,

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
128 = $${ 2 }^{ 7 }$$

(ii) We have,

243 = 3 x 3 x 3 x 3 x 3
243 = $${ 3 }^{ 5 }$$

(iii) We have,

625 = 5 x 5 x 5 x 5

625 = $${ 5 }^{ 4 }$$

## Positive Integral Exponent of a Rational Number

Let $$\frac{a}{b}$$ be any rational number and n be a positive integer. Then,

$${(\frac { a } { b })^{ n } }$$ = $$\frac{a}{b}$$ x $$\frac{a}{b}$$ x $$\frac{a}{b}$$…n times
= $$\frac { a\quad \times \quad a\quad \times \quad a….n\quad times }{ b\quad \times \quad b\quad \times \quad b….n\quad times }$$
= $$\frac { { a }^{ n } }{ { b }^{ n } }$$
Thus, $${(\frac { a }{ b })^{ n } }$$ = $$\frac { { a }^{ n } }{ { b }^{ n } }$$ for every positive integer n.

Example : Evaluate:

(i) $${(\frac { 3 } { 7 })^{ 3 } }$$    (ii) $${(\frac { -2 }{ 5 })^{ 3 } }$$

Solution.

(i) $${(\frac { 3 } { 7 })^{ 3 } }$$ = $$\frac { { 3 }^{ 3 } }{ { 7 }^{ 3 } }$$ = $$\frac{127}{343}$$

(ii) $${(\frac { -2 } { 5 }) ^{ 3 } }$$ = $$\frac { { (-2) }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$-\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$-\frac{8}{125}$$

## Negative Integral Exponent of a Rational Number

Let $$\frac{a}{b}$$ be any rational number and n be a positive integer.

Then, we define,  $${(\frac { a }{ b })^{ -n } }$$ = $${(\frac { b }{ a })^{ n } }$$

Example : Evaluate:

(i) $${(\frac { 1 } { 2 })^{ -3 } }$$    (ii) $${(\frac { 2 } { 7 })^{ -2 } }$$

Solution.

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

(ii) $${(\frac { 2 } { 7 })^{ -2 } }$$
= $${(\frac { 7 } { 2 })^{ 2 } }$$

= $$\frac { { 7}^{ 2 } }{ { 2 }^{ 2 } }$$
= $$\frac{49}{4}$$