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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 4th 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 xm *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)
=a18 *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 xm /xn = x(m-n)
If x is a non zero real number and m and n are numbers
with m<n then xm /xn = 1/(x(n-m) )
By definition, for any x 0
1) xm
= 1/( x-m)
2) x-m
= 1/ ( xm)
3) If n is
a positive integer and a 0 then = a1/n
4) If
a 0 and n 0 then am/n=
Observe:
x0 =1 (1 = xm /xm = 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 xa+b /xb-c
Solution:
xa+b /xb-c
= xa+b /1/(x-(b-c)) (By definition)
= xa+b *x-(b-c)
= xa+b+(-(b-c)) (By Second law)
= xa+b-b+c(-(b-c) = -b+c)
= xa+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
= 52*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 (xm )n = xmn
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 = (xm)* (ym)
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 --ŕBy fourth law
= (3) -1* x+2 *y-1 ----ŕ By third law
= x2 /3*y--ŕ 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 = (xm)/ (ym)
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 *am =
am
2. If m is an odd number then (-a)m == (-1)m *am = -am
Proof :
1.
If
m is even then m is of the form 2n ( n=1,2,3..)
(-1)m = (-1)2n = ((-1)2 )n ----ŕ 3rd 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
----ŕ 2nd Law.
= 1n *-1 ----ŕ(proved in the previous case )
= -1
2.2
Problem 8 :
Simplify (am/an)p*(an/ap)m*(ap/am)n
Solution :
(am/an)p
= (am)p/(an)p (By fifth law)
= amp/ anp
(By third law)
(am/an)p*(an/ap)m*(ap/am)n
= (amp/ anp)* (anm/ apm)*
(apn/ amn) (By expanding other terms)
= (amp* anm* apn)/ (anp*apm*amn)(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) = (a4-2) = 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/( xm)
=(a-12+6)* (b15-12) ( 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 xp/q
as the ‘qth root’ of xp or
alternatively
xp/q as the ‘pth
power ’ of x1/q.
i.e. xp/q =
(xp)1/q=
= (x1/q)p= ( )p
Note: (r/s)-p/q =(s/r)p/q
The form x1/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 = 2a5b7c calculate the value of = 2-a7b5-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-a7b5-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
3x-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.