Today our topic of discussion is Exponents or Indices.
Exponents or Indices
Exponents and Logarithms
Very large or very small numbers or expressions can easily be expressed by using exponents. As a result, calculations and solution of mathematical problems become easier. Scientific or standard form of a number is expressed by using exponents. Therefore, every student should have the knowledge of exponents and its applications.
Exponents be get logarithms. Multiplication and division of numbers or expressions and exponent related calculations have become easier with the help of logarithms. Use of logarithm in scientific calculations was the only way before calculators and computers became available. Still the use of logarithm is important as the alternative of calculator and computer.
In this chapter, exponents and logarithms have been discussed in detail.
At the end of the chapter, the students will be able to –
‣ explain the rational exponent.
‣ explain and apply the positive integral exponents, zero and negative integral exponents.
‣ solve problems by describing and applying the rules of exponents.
‣ explain the n th root and rational fractional exponents and express the n th root in terms of exponents.
▸ explain logarithms.
‣ prove and apply the formulae of logarithms.
‣ explain natural logarithm and common logarithm.
‣ explain the scientific form of numbers.
‣ explain the characteristic and mantissa of common logarithm.
‣ calculate common and natural logarithm by calculator.
Exponents or Indices
In class VI, we have got the idea of exponents and in class VII, we have learnt the exponential rules for multiplication and division. Expressions associated with exponent and base are called exponential expression.
If a is any real number, successive multiplication of n as is an that is, aa * a *…* a (n times a) = a”, where n is a positive integer. Here, n is index or power and a is base. Conversely a” = axaxax…xa (n times a).
Exponents may not only be positive integers, they may be negative integers or positive fractions or negative fractions. That is, for base a in R (set of real numbers) and power n in Q (set of rational numbers), an is defined.
The case of n in N (set of natural numbers) is especially considered. Besides, exponents may also be irrational. But as it is out of curriculum, it has not been discussed in this chapter.
Index Laws
Let, a in R (set of real numbers) and m ,n in N (set of natural numbers).
Formula 1 (Multiplication). am* an = a(m+n)
Formula 2 (Division). (am)/(an) = a(m-n) * when m ≤ n; 1/(a(n-m) * when n > m
∴ In general am * an = a(m+n) and (am)/(an) = a(m-n) * whenm >= n; 1/(a(n-m) )* whenn > m Formula 3 (power of product). (ab)n = an * bn
We observe, (5 x 2)3 = (5 × 2) x (5 × 2) × (5 × 2) [a³ = axaxa, a = 5 * 2 ] = (5 * 5 * 5)(2 * 2 * 2) = 53 * 23
In general, (ab)” ab x ab x ab x x ab [successive multiplication of n ab’s] = =( aaa *…* a)*(b* b* b*…* b) = an * bn
Formula 4 (power of quotient).
We observe,
In general,
(a/b)n = (an)/(bn)
(5/2)3 = 5/2 * 5/2 * 5/2 = (53)/(23)
( a b )n = a/b * a/b * a/b …* a b [successive multiplication of 7 times = a* a* a*…* a b* b* b*…* b = (an)/(bn)
Formula 5 (power of power). (am)n = amn
(am)n = am * am * a m *…* am successive multiplication of n times a m ] =am + m +m…+m [in the power, multiplication of n times of exponent] = a mn = a mn
.. (a m)n = amn
Zero and Negative Indices
For 0 and negative indices a0 and a(-n) are defined as follows (where n is natural number).
Definition 1 (zero index). a0 = 1 , ( a ne 0)
Definition 2 (negative index). a(-n)= 1/(an) ( a ne 0,n in N)
For all integer indices m and n the definition Observe, (an)/(an) = a p(n – n) = a0
But axaxax…xa (n times) axaxax…xa (n times) 1
a0 = 1
and 1/(an) = (a0)/(an) = a (0-n) = a(-n)
Example 1. Find the values:
1) (52)/(53)
2) (2/3)5 * (2/3)-5
Solution:
1) (52)/(53) = 5(2-3) = 5-1 = 1/(51) = 1/5
(2/3)5 * (2/3)-5= (2/3) (5 – 5) = (2/3)0 = 1 2)
Example 2. Simplify:
1)(54 * 8 * 16)/(25 * 125)
2) (3 * 2n-4 * 2(n-2)/2(n-2)(n-1)
Solution:
1) (54 * 8 * 16)/(25* 125) = (54 * 23 * 24)/(25 * 53)
= 54 * 2(3 + 4))/(53 * 25) ^ (3 + 4))/(53 * 25)
= (54)/(53) * (27)/(25)= 54-3 * 2(7-3) = 51 * 22 = 5 * 4 = 20
2) (3 * 2 ^ n – 4 * 2 ^ (n – 2))/(2n-1 ^ (n – 1))
= (3 * 2 ^ n – 2 ^ 2 * 2 ^ (n – 2))/(2 ^ n – 2 ^ n * 2-1)
= (3 * 2 ^ n – 2 ^ (2 + n – 2))/(2 ^ n – 2 ^ n * 1/2)
= (3 * 2 ^ n – 2n)/((1 – 1/2) * 2n)
= ((3 – 1) * 2n)/(1/2 * 2n) = (2 * 2n)/(1/2 * 2n) = 2 * 2 = 4
Example 3. Prove that, (ap) ^ (q – tau) * (aq) ^ (r – p) * (a ^ r) ^ (p – q) = 1
Solution: (ap) ^ (q – r) * (a ^ q) ^ (r – p) * (ar) ^ (p – q)
=a^ p(q-r) * a^ q(r-p) * a^ r(p-q) [ vdots(a^ m )^ n =amn ] = a ^ (pq – pr) * a ^ (qr – pq) * a ^ (pr – qr) = a ^ (pq – pr + qr – pq + pr – qr) = a0= 1
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