2.2 Exponents:

How many zeros are there in a crore and in thousand crores?

Observe the following few quotes from Yuddha kanda in Ramayana . It’s period is said to be at least 4000BC.

Shatam shata sahasranam koti mahurmahishana: ||1||

Meaning: 100 * 100 * 1000 = koti(crore)

Shatam koti sahasranam shanka ityabhidhiyate ||2||

Meaning: 100 * koti * 1000 = Shankha

Shatam shankha sahasranam mahashankha itismrata: || 3||

Meaning: 100 * shankha * 1000 = Mahashankha

The verses go on till Mahougha which is 1 followed by 62 zeros (= 10⁶²)

This shows that during those days itself, they were familiar with zero and well conversant with the decimal number system.

In this topic we learn easy method of representation and multiplication of large numbers

We know 16 = 2*2*2*2 (the number 2 is multiplied 4 times)

Therefore we say 16 is equal to 4^th power of 2 and write 16 = 2⁴. We say 16 is equal to ‘2 raised to the power of 4’

16 = 4*4 = 4² (We also say 16 is equal to ‘4 raised to the power of 2’)

Like the way we factorize numbers we can also factorize algebraic expressions.

For example x³= x*x*x

We say x³ is ‘x raised to the power of 3’

This method of easy representation of x*x*x by x³ is called ‘exponential notation’.

In general:

xⁿ = x *x*x* …. n times

Here x is called base and n is called exponent or ‘index’

Base ^Exponent = Number

Note a = a¹

2.2 Problem 1: Express 1331 to base of 11

Solution:

We know the factors of 1331 are 11, 11, 11

1331 = 11*11*11 = 11³

Let us find the product of 2⁵ and 2³

2⁵ *2³ = (2*2*2*2*2)*(2*2*2) = 2⁸

Did you notice that 8 =5+3?

1. From the above observation we arrive at the first law (Product Law) of exponents:

If x is a non zero real number and m and n are numbers then x^m *xⁿ = x^(m+n)

2.2 Problem 2: Evaluate a¹⁴ *b³² * a⁴ *b¹⁶

Solution:

a¹⁴ *b³² * a⁴ *b¹⁶

= (a¹⁴ * a⁴ )*(b³² * b¹⁶) ( By rearranging terms)

= (a¹⁴⁺⁴⁾*(b³²⁺¹⁶) (By first law)

=a^{18 *}b⁴⁸

Let us divide 2⁵ by 2³

2⁵ /2³ = (2*2*2*2*2)/(2*2*2) = 2*2=2²

Similarly 2³ /2⁵ = (2*2*2)/ (2*2*2*2*2) = 1/(2*2) = 1/(2²)

2³ /2³ = (2*2*2)/(2*2*2) = 1

2. From the above observation we arrive at the second law (Quotient Law) of exponents:

If x is a non zero real number and m and n are numbers with m>n then x^m /xⁿ = x^(m-n)

If x is a non zero real number and m and n are numbers with m<n then x^m /xⁿ = 1/(x^(n-m) )

By definition, for any x 0

1) x^m = 1/( x-^m)

2) x^-m = 1/ ( x^m)

3) If n is a positive integer and a 0 then = a^1/n

4) If a 0 and n 0 then a^m/n=

Observe:

x⁰ =1 (1 = x^m /x^m = x^(m-m) )

2.2 Problem 3: Write equivalents for 10^{-5 and} 2/m^-1

Solution:

10^-5 = 1/10⁵

2/m^-1= 2/(1/m¹) = 2m¹ =2m

2.2 Problem 4: Evaluate x^a+b /x^b-c

Solution:

x^a+b /x^b-c

= x^a+b /1/(x^-(b-c)) (By definition)

= x^a+b *x^-(b-c)

= x^a+b+(-(b-c)) (By Second law)

= x^a+b-b+c(-(b-c) = -b+c)

= x^a+c

Let us find the product of 5² , 5² and 5²

5² *5²*5²= (5*5)*(5*5)*(5*5) = 5⁶

We can write this also as

5² *5²*5² = (5²)³ = 5^2*3

3. From the above observation we arrive at the third law (Power law) of exponents:

If x is a non zero real number and m and n are numbers then (x^m )ⁿ = x^mn

2.2 Problem 5 : Evaluate [{(x²)²}²]²

Solution:

(x²)²= x⁴

{(x²)²}² = {x⁴}² = x⁸

[{(x²)²}²]² = [x⁸]²= x¹⁶

Exercise: Verify the answer by expanding the terms individually

Let us evaluate (2*5)³

(2*5)³ = (2*5)*(2*5)*(2*5) (By definition)

= (2*2*2)*(5*5*5) (By Grouping all 2 and 5 together)

= (2)³*(5)³

4. From the above observation we arrive at the fourth law of exponents:

If x and y are non zero real numbers and m is a number then (x*y)^m = (x^m)* (y^m)

2.2 Problem 6: Simplify (5x^-3 y^-2)³

Solution:

(5x^-3 y^-2)³

= (5)³ *(x^-3)³*(y^-2)³ ( By fourth law)

= 5³* x^-9* y^-6 (By third law)

= 5³/( x⁹* y⁶) (By definition)

Exercise: Verify the answer by expanding the terms individually

2.2 Problem 7: Simplify (3x^-2 y)^-1

Solution:

(3x^-2 y)^-1

= (3) ^-1*( x^-2)^-1 *(y)^-1 --àBy fourth law

= (3) ^-1* x⁺² *y^-1 ----à By third law

= x² /3*y--à By definition

Verify the answer by expanding the terms individually

Let us evaluate (2*5)³

(2/5)³ = (2/5)*(2/5)*(2/5) (By definition)

= (2*2*2)/(5*5*5) ( Group all 2 and 5 together)

= (2)³/(5)³

4. From the above observation we arrive at the fifth law of exponents:

If x and y are non zero real numbers and m is a number then (x/y)^m = (x^m)/ (y^m)

Observations:

Since (-1)² = (-1)*(-1) =+1 and (-1)³ = (-1)*(-1)*(-1) = -1 we note:

1. If m is an even number then (-a)^m = (-1)^m *a^m = a^m

2. If m is an odd number then (-a)^m == (-1)^m *a^m = -a^m

Proof :

1. If m is even then m is of the form 2n ( n=1,2,3..)

(-1)^m = (-1)²ⁿ = ((-1)² )ⁿ ----à 3^rd law

= 1ⁿ = 1

2. If m is odd then m is of the form 2n+1 ( n=0,1,2,3..)

(-1)^m = (-1)²ⁿ⁺¹ = (-1)²ⁿ *(-1)¹ ----à 2^nd Law.

= 1ⁿ *-1 ----à(proved in the previous case )

= -1

2.2 Problem 8 : Simplify (a^m/aⁿ)^p*(aⁿ/a^p)^m*(a^p/a^m)ⁿ

Solution :

(a^m/aⁿ)^p

= (a^m)^p/(aⁿ)^p (By fifth law)

= a^mp/ a^np (By third law)

(a^m/aⁿ)^p*(aⁿ/a^p)^m*(a^p/a^m)ⁿ

= (a^mp/ a^np)* (a^nm/ a^pm)* (a^pn/ a^mn) (By expanding other terms)

= (a^mp* a^nm* a^pn)/ (a^np*a^pm*a^mn)(Note numerator and denominator are same)

=1

2.2 Problem 9 : Simplify (a⁴b^-5/ a²b^-4)^-3

Solution:

Let us first, simplify the term

(a⁴b^-5/ a²b^-4)

= (a⁴/ a²) * (b^-5/ b^-4)

= (a²/ b) ( By second law - (a⁴/ a²) = (a^4-2) = a², (b^-5/ b^-4) = (b^-5-(-4)) = b^-5+4= b^-1= 1/b )

Now let us take the given problem

(a⁴b^-5/ a²b^-4)^-3

= (a²/ b)^-3 (Substitute the simplified term for (a⁴b^-5/ a²b^-4)

= (a²)^-3/ (b)^-3 (By third law)

=a^-6/b^-3

= b³/a⁶

Alternate method: Let us solve this problem in another way

(a⁴b^-5/ a²b^-4)^-3

= (a^-12b⁺¹⁵/ a^-6b⁺¹²) ^{(By third law)}

=(a^-12/ a^-6)* (b¹⁵/ b¹²) ( By grouping terms)

=(a^-12* a⁶)* (b¹⁵* b^{- 12}) ( By definition x ^-m = 1/( x^m)

=(a^-12+6)* (b^15-12) ( By first law)

=a^-6*b³

= b³/a⁶

2.2 Summary of learning

No	Points to remember
1	By definition x^n= x*x*x*x – n times
2	Base ^Exponent= Number
3	By definition x^0=1
4	By definition x ^- m= 1/( x^m)
5	First law :x^m*x^n = x^(m+n)
6	Second law x^m/x^n = x^(m-n)
7	Third law (x^m)^n = x^mn
8	Fourth law (x*y)^m = (x^m)* (y^m)
9	Fifth law (x/y)^m = (x^m)/ (y^m)

Additional Points:

If x is a positive rational number and m(=p/q) a positive rational exponent, then we define x^p/q as the ‘qth root’ of x^p or alternatively

x^p/q as the ‘pth power ’ of x^1/q.

i.e. x^p/q = (x^p)^1/q= = (x^1/q)^p= ( )^p

Note: (r/s)^-p/q =(s/r)^p/q

The form x^{1/m is} called ‘exponential form’ and if m>0, then the form is called ‘radical form’. The sign is called the radical sign and is called ‘radical’. The number m is called ‘index’ of the radical and x is called the ‘radicand’.

Note, index is always a positive number .

2.2 Problem 10: If 1960 = 2^a5^b7^c calculate the value of = 2^-a7^b5^-c

Solution:

1960 = 2*2*2*5*7*7= 2³5¹7²

a=3, b=1 and c=2

Hence

2^-a =1/8 and 5^-c =1/25

Thus

2^-a7^b5^-c

= (1/8)*7*(1/25) = 7/200

2.2 Problem 11: Simplify {(8x³)/125y³}^2/3

Solution:

We know:

8x³ = (2x)³ and 125y³ = (5y)³

(8x³)/125y³

= (2x/5y)³

{(8x³)/125y³}^2/3

= {(2x/5y)³}^2/3

= (2x/5y)³*^2/3

= (2x/5y)²

= 4x²/25y²

2.2 Problem 12: Find x if 3x^-1 = 9*3⁴

Solution:

9 = 3²

9*3⁴ = 3²*3⁴ = 3⁶

Since

3^x-1 = 9*3⁴ = 3⁶,

x-1 = 6

x=7

Verification:

Substitute x=7 in the given problem and expand the terms to note that the answer = 729.