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Indices |
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Large or small numbers are better expressed in terms of indices. A given number is written as a base raised to the index, that is (base)index. Bases and indices can be any real number. In the last chapter we have seen what are squares and cubes of numbers; for squares the index is 2, for cubes the index is 3. For square roots the index is (1/2) and for cube roots the index is (1/3). The base has to be a suitable number that together with the indices gives the correct number. If the index is n, then the resultant number is obtained by multiplying the base n times. Laws of indicesWe will state a few facts
about indices and try to see their validity using all sorts of numbers. The
laws are valid for all real numbers, but for the present syllabus, it is
sufficient to consider only rational numbers. 1. am * an = a (m+n). 2. a (-m) = 1 / am 3.
am / an 4. (am) n = a (m * n). 5.
(a * b) m = am * bm 6.
(a / b) m = am / bm 7.
a0 = 1 1. To prove that am * an = a (m+n). Example
1 : Let us consider the base
a = 2, let the value of indices m and n be m=4, n=5 LHS
: am * an = 24 * 25 = (2 * 2 * 2
* 2) * (2 * 2 * 2 * 2 * 2) = 16 * 32 = 512 RHS
: a (m+n) = 29
= 512 Since
LHS = RHS, hence it is proved that am * an = a (m+n) Example
2 : Let us consider the
base a = 10, let the value of indices m and n be m=2, n=3 LHS
: am * an = 102 * 103 = (10 * 10)
* (10 * 10 * 10) = 100 * 1000 = 100000 RHS
: a (m+n) = 105
= 100000 Since
LHS = RHS, hence it is proved that am * an = a (m+n) 2. To prove that a (-m) = 1 / am. Multiply
both the LHS and the RHS by am
a (-m)
* am = (1 / am) From
first law of indices proved above, LHS becomes a (m-m) = a0 In
the RHS, the numerator and the denominator cancel each other out to give a
resultant of 1. Thus
a (m-m) = 1, which means that a (-m) * am = 1. This gives a (-m) =
1/ am 3. From the above two proofs, the third law of indices am / an = a (m-n)
, a can
be easily deduced. The
condition a If
m > n, the index of the expression on the RHS, that is (m-n), will be a
positive number. If
m < n, the index of the expression on the RHS, the is (m-n), will be
a negative number. In this case the second law of indices will have to be
applied to obtain the value. 4.
To prove that (am) n =
a (m * n) Example
1 : Let us consider the
base a = 7, let the value of indices m and n be m=2, n=5. LHS
: (am) n = (72) 5 = ( 7 * 7 )
5 = 495
= 49 * 49 * 49 * 49 * 49 = 282475249 RHS
: a (m * n) = 710 =
7 * 7 * 7 * 7 * 7 * 7 * 7 * 7 * 7 * 7
= 282475249 Thus,
LHS = RHS, hence it is proved that (am) n = a (m *
n) Example
2 : Let us consider the
base a = 1/2, let the value of indices m and n be m=1, n=2. LHS
: (am) n
= ( (1/2)1 )2 =
(1/2)2 = ‡ *
‡ = º RHS
: a (m * n) = (1/2)2
= ‡ * ‡ = º Thus
LHS = RHS, hence it is proved that (am) n =
a (m * n). 5.
To prove that (a * b) m
= am * bm.
Example
1 : Let us consider the
base a = 2 and b = 3, let the value of index m be m = 2 LHS
: (a * b) m = ( 2 * 3 )2 = 62 = 6 * 6 = 36 RHS
: am * bm =
22 * 32 = 4 * 9 = 36 Thus
LHS = RHS, hence it is proved that (a * b) m = am * bm.
Example
2 : Let us consider the
base a = 4 and b = 9, let the value of index m be m = ‡ LHS
: (a * b) m = ( 4 * 9 )1/2 = 361/2
= 6 RHS
: am * bm =
41/2 * 91/2
= 2 * 3 = 6. Thus
LHS = RHS, hence it is proved that (a * b) m = am * bm.
6. From the 1st , 2nd and the 5th laws of indices the 6th law of indices can also be proved. Thus,
(a/b)m =
am /
bm , b The
condition b 7. The
7th law of indices a0 = 1 is already proved above
in the 2nd law of
indices. Summary
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