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 (= 10^{62})
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 = 2^{4}.
We say 16 is equal to ‘2 raised to the power of 4’
16 = 4*4 = 4^{2 }(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 x^{3}=
x*x*x
We say x^{3
} is ‘x raised to the power
of 3’
This method of easy
representation of x*x*x by x^{3 }is called ‘exponential
notation’.
In general:
x^{n}^{ } = x *x*x* …. n times
Here x is called base and n is called exponent
or ‘index’
Base ^{Exponent }= Number
Note
a = a^{1 }
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 = 11^{3}
Let us find the product of 2^{5} and 2^{3}^{}
_{} 2^{5 }*2^{3}
= (2*2*2*2*2)*(2*2*2) = 2^{8}
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}^{ }*x^{n}^{
} = x^{(m+n) }
2.2 Problem 2: Evaluate a^{14 }*b^{32} * a^{4 }*b^{16} ^{}
Solution:
a^{14 }*b^{32}
* a^{4 }*b^{16}
= (a^{14 }* a^{4 })*(b^{32} * b^{16}) ( By rearranging terms)
= (a^{14+4)}*(b^{32+16})
(By first law)
=a^{18 *}b^{48}
Let us divide 2^{5} by 2^{3}
_{} 2^{5 }/2^{3} = (2*2*2*2*2)/(2*2*2) = 2*2=2^{2}
Similarly 2^{3 }/2^{5 }=
(2*2*2)/ (2*2*2*2*2) = 1/(2*2) = 1/(2^{2})
2^{3 }/2^{3 }=
(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}^{ }/x^{n}^{
} = x^{(m-n) }
If x is a non zero real
number and m and n are numbers with m<n then x^{m}^{ }/x^{n}^{
} = 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:
x^{0}^{ }=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 }= 1/10^{5}
2/m^{-1}= 2/(1/m^{1}) = 2m^{1} =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 5^{2}
, 5^{2 }and 5^{2}
_{} 5^{2 }*5^{2}*5^{2}=
(5*5)*(5*5)*(5*5) = 5^{6}
We can write this also as
5^{2 }*5^{2}*5^{2}
= (5^{2})^{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 [{(x^{2})^{2}}^{2}]^{2}
Solution:
(x^{2})^{2}= x^{4}
{(x^{2})^{2}}^{2} = {x^{4}}^{2} = x^{8}
[{(x^{2})^{2}}^{2}]^{2} = [x^{8}]^{2}= x^{16}
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)
= 5^{3}* x^{-9}*
y^{-6 } (By third law)
= 5^{3}/( x^{9}* y^{6}) (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
= x^{2} /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}^{ } = (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 }----ŕ 3^{rd}
law
= 1^{n }= 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 2^{nd} Law.
= 1^{n }*-1 ^{ }----ŕ(proved in
the previous case )
= -1^{}
2.2 Problem 8 :
Simplify (a^{m}/a^{n})^{p}*(a^{n}/a^{p})^{m}*(a^{p}/a^{m})^{n}
Solution
:
(a^{m}/a^{n})^{p}
= (a^{m})^{p}/(a^{n})^{p} (By fifth law)
= a^{mp}/ a^{np}^{ } (By third law)
_{} (a^{m}/a^{n})^{p}*(a^{n}/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 (a^{4}b^{-5}/ a^{2}b^{-4})^{-3}
Solution:
Let
us first, simplify the term
(a^{4}b^{-5}/
a^{2}b^{-4})
= (a^{4}/ a^{2})
* (b^{-5}/ b^{-4})
= (a^{2}/ b) (_{} By second law - (a^{4}/
a^{2}) = (a^{4-2}) = a^{2},
(b^{-5}/ b^{-4}) = (b^{-5-(-4)}) = b^{-5+4}= b^{-1}=
1/b
)
Now let us take the given
problem
(a^{4}b^{-5}/
a^{2}b^{-4})^{-3}
= (a^{2}/ b)^{-3 }(Substitute the simplified term
for (a^{4}b^{-5}/ a^{2}b^{-4})
= (a^{2})^{-3}/
(b)^{-3} (By third law)
=a^{-6}/b^{-3}
= b^{3}/a^{6}
Alternate
method: Let us solve this problem in another way
(a^{4}b^{-5}/ a^{2}b^{-4})^{-3}
= (a^{-12}b^{+15}/ a^{-6}b^{+12})^{ (By third law)}
=(a^{-12}/
a^{-6})* (b^{15}/ b^{12}) ( By grouping terms)
=(a^{-12}*
a^{6})* (b^{15}* b^{- 12}) ( By definition x ^{-m }=
1/( x^{m})
=(a^{-12+6})*
(b^{15-12}) ( By
first law)
=a^{-6}*b^{3}
= b^{3}/a^{6}
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^{a}5^{b}7^{c} calculate the value of = 2^{-a}7^{b}5^{-c}
Solution:
1960 = 2*2*2*5*7*7=
2^{3}5^{1}7^{2}
_{} a=3, b=1 and c=2
Hence
2^{-a }=1/8 and 5^{-c }=1/25
Thus
2^{-a}7^{b}5^{-c}
= (1/8)*7*(1/25) = 7/200
2.2 Problem 11:
Simplify {(8x^{3})/125y^{3}}^{2/3}
Solution:
We know:
8x^{3 }= (2x)^{3} and 125y^{3 }= (5y)^{3}
_{} (8x^{3})/125y^{3}
= (2x/5y)^{3}
_{}{(8x^{3})/125y^{3}}^{2/3}
= {(2x/5y)^{3}}^{2/3}
= (2x/5y)^{3}*^{2/3}
= (2x/5y)^{2}^{}
= 4x^{2}/25y^{2}^{}
^{ }
2.2 Problem 12:
Find x if 3x^{-1} = 9*3^{4}
Solution:
9 = 3^{2}
_{} 9*3^{4 }= 3^{2}*3^{4
}= 3^{6}
Since
3^{x-1} = 9*3^{4}
= 3^{6},
x-1
= 6
_{} x=7
Verification:
Substitute x=7 in the given
problem and expand the terms to note that the answer = 729.