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 xn = x*x*x*x – n times
2	Base Exponent = Number
3	By definition  x0 =1
4	By definition   x - m = 1/( xm)
5	First law :xm  *xn   = x(m+n)
6	Second law xm  /xn   = x(m-n)
7	Third law (xm  )n   = xmn
8	Fourth law (x*y)m    = (xm)* (ym)
9	Fifth law (x/y)m    = (xm)/ (ym)

 

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.