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8 lessons,
56m 1s
Concept of Logarithm
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This lesson will help you understand the basic concept of Logarithms

Abhishek Khurana
I am Abhishek Khurana. I completed my B.Com(H) from Panjab University in 2008 & my MBA from NMIMS in 2011.

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  1. Abhishek Khurana 6 times 99 percentiler in CAT MBA from NMIMS 2009-2011 8 years of teaching experience


  3. Introduction 1.Every positive real number N can be expressed in exponential form as N a.1) e.g. 49 72 where a' is also a positive real different than unity and is called the base and 'x is called the exponent. 2. We can write the relation (1) in logarithmic form as log,N x... Hence the two relations ax-N and LogaN-X are identical where N0a0a not equal to zero. Hence logarithm of a number to some base is the exponent by which the base must be raised in order to get that number. Logarithm of zero does not exist and logarithm of Hive reals are not defined in the system of real numbers.

  4. Introduction EXAMPLE 1: Find value of (0) logs127 ( log1o100

  5. Introduction EXAMPLE 1: Find value of (0) log8127 logo100 Solution: (i) Let log8127 = x 27 = 814 33 34x So x 3/4 i Let log1o100 x 100= 10% 102 =10x gives x = 2

  6. Introduction (a) Unity has been excluded from the base of the logarithm as in this case log,N will not be possible and if N 1 then log, 1 will have infinitely many solutions and will not be unique which is necessary in the functional notation. b) alog N N is an identify for all N> 0 and a 0,anot equal to 1 e.g. 5092 2-5 c) Using the basic definition of log we have 3 important deductions: D logNN-1 i.e. logarithm of a number to the same base is 1. ) logiN11 i.e. logarithm of a number to its reciprocal is -1. m) loga 10i.e. logarithm of unity to any base is zero.

  7. Properties of Logarithm, n are arbitrary positive real numbers where a > 0 ; a #1 logam + loga n = logam n (m > 0, n > 0) Proof: Let xl - logom m-a* x2=loga n ; n =ax2 Now mn ax xl t x2log mn logom + logon logomn 2.loga m/n = logam-logan

  8. Properties of Logarithm EXAMPLE 2 Find the value of x satisfying logo (24 x-41) = x (1-log,05) Solution: logo (2x + x-41)=x(1-log105) log10(2x+x-41) = x log102. logo (29 2x + X-41 = 2x x = 41

  9. Properties of Logarithm EXAMPLE 3 For 0 < at 1, find the number of ordered pair (x, y) satisfying the equation logflxty|- and log ay-log al x|-log a24, Solution: we have log a2 Ixtyl = => |x+yl=a => x+y= a Also, loga (y/lx/)=loga24 => y=21x/ (2) If x > 0, then x a/3 , y = 2a/3 if x < 0, then y = 2a, x=-a possible ordered pairs (a/3,2a/3) and -a, 2a) (1)