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^{(mn)
}
If x is a non zero real number and m and n are numbers
with m<n then x^{m }/x^{n } = 1/(x^{(nm) })
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^{(mm)
})
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^{bc} ^{}
Solution:
x^{a+b }/x^{bc}
= x^{a+b }/1/(x^{(bc)}) (By
definition)
= x^{a+b }*x^{(bc)}
= x^{a+b+((bc)) }(By Second law)
= x^{a+bb+c}(_{}(bc) = 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*yŕ 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^{42}) = 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^{1512}) ( 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^{(mn) } 
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^{x1} = 9*3^{4} = 3^{6},
x1 = 6
_{} x=7
Verification:
Substitute x=7 in the given problem and expand the
terms to note that the answer = 729.