Logarithms Chapter 13.1 PDF

E3 60 2 Logarithms 1.1.1 Definition. “The Logarithm of a given number to a given base is the index of the power to which the base must be ID raised in order to equal the given number.” If a  0 and  1, then logarithm of a positive number N is defined as the index x of that power of 'a' which U equa...

E3 60 2 Logarithms 1.1.1 Definition. “The Logarithm of a given number to a given base is the index of the power to which the base must be ID raised in order to equal the given number.” If a  0 and  1, then logarithm of a positive number N is defined as the index x of that power of 'a' which U equals N i.e., log a N  x iff a x  N  a log a N  N, a  0 , a  1 and N  0 It is also known as fundamental logarithmic identity. D YG The function defined by f (x )  log a x , a  0, a  1 is called logarithmic function. Its domain is (0, ) and range is R. a is called the base of the logarithmic function. When base is 'e' then the logarithmic function is called natural or Napierian logarithmic function and when base is 10, then it is called common logarithmic function. Note :  The logarithm of a number is unique i.e. No number can have two different log to a given base. U  log e a  log e 10. log 10 a or log 10 a  log e a  0.434 log e a log e 10 ST 1.1.2 Characteristic and Mantissa. (1) The integral part of a logarithm is called the characteristic and the fractional part is called mantissa. log 10 N  integer  Characters tics  fraction (ve )  Mantissa (2) The mantissa part of log of a number is always kept positive. (3) If the characteristics of log 10 N be n, then the number of digits in N is (n+1) (4) If the characteristics of log 10 N be (– n) then there exists (n – 1) number of zeros after decimal part of N. Logarithms 3 For y  log a x to be defined 'a' must be Solution: (d) (a) Any positive real number (b) Any number (c) (d) Any positive real number  1 e It is obvious (Definition). Logarithm of 32 5 4 to the base 2 2 is (a) 3.6 (c) 5.6 Let x be the required logarithm , then by definition (2 2 )x  32 5 4 (2.2 1 / 2 ) x  2 5.2 2 / 5 ;  2 3x 2 2 Here, by equating the indices, 5 2 5 3 18 27 x  3.6 , x  2 5 5 (d) None of these E3 Solution: (a) (b) 5 ID Example: 2 [IIT 1990] 60 Example: 1 1.1.3 Properties of Logarithms. (1) log a a  1, log a 1  0 U Let m and n be arbitrary positive numbers such that a  0, a  1, b  0, b  1 then (2) log a b. log b a  1  log a a  log b b  log a b  log b a log b c (4) log a (mn )  log a m  log a n D YG (3) log c a  log b a. log c b or log c a  m  (5) log a    log a m  log a n n (6) log a m n  n log a m (7) a loga m  m 1 (8) log a     log a n n 1  (10) log a  n   log a n U (9) log a  n  Example: 3 The number log 2 7 is (a) An integer Solution: (c)  log a n , (  0 )  (a, b, c  0 and c  1) ST (11) a logc b  b logc a , 1 log b a [IIT 1990] (b) A rational number (c) An irrational number (d) A prime number Suppose, if possible, log 2 7 is rational, say p / q where p and q are integers, prime to each other. Then, p  log 2 7 q  7  2 p / q  2 p  7q , Which is false since L.H.S is even and R.H.S is odd. Obviously log 2 7 is not an integer and hence not a prime number Example: 4 If log 7 2  m, then log 49 28 is equal to [Roorkee 1999] 4 Logarithms 1  2m 2 (b) 2 (1  2m) (c) 2 1  2m 1 1 1 1 1  2m log 7 log 4 log 28 log 7  log 4 1 1 =    log 7 4 = .2 log 7 2 =  log 7 2   m   2 2 2 2 2 2 log 7 2 log 7 2 2 log 49 2 log 7 Solution: (b) log 49 28  Example: 5 ab  1 If log e    (log e a  log e b) , then relation between a and b will be  2  2 Solution: (a) b 2 (c) 1 ab  1 log e    (log e a  log e b)  log e (ab)  log e 2  2  2   a  b ab  ab  a  b  2 ab  2 2 Solution: (c) (b) 19 Let y  3 40 is b 3 0  a  b 0  ab If log 10 3  0.477 , the number of digits in 3 40 is (a) 18 ab (d) a  2a  b ID Example: 6 (b) a  ab [UPSEAT 2000] E3 (a) (d) 1  m 60 (a) (c) 20 [IIT 1992] (d) 21  Number of digits in y  19  1  20 Which is the correct order for a given number  in increasing order (a) (c) Solution: (b) D YG Example: 7 U Taking log both the sides, log y  log 3 40  log y  40 log 3  log y  19.08 [Roorkee 2000] log 2 , log 3 , log e , log10  (b) log10 , log 3 , log e , log 2  log10 , log e , log 2 , log 3  (d) log 3 , log e , log 2 , log10  Since 10, 3, e, 2 are in decreasing order U Obviously, log10 , log 3 , log e , log 2  are in increasing order. 1.1.4 Logarithmic Inequalities. (2) If 0  a  1, p  1  log a p  0 (3) If a  1, 0  p  1  log a p  0 (4) If p  a  1  log a p  1 (5) If a  p  1  0  log a p  1 (6) If 0  a  p  1  0  log a p  1 (7) If 0  p  a  1  log a p  1 a  m b , if m  1 (8) If log m a  b   b ST (1) If a  1, p  1  log a p  0 a  m b , if m  1 (9) log m a  b   b  a  m , if 0  m  1  a  m , if 0  m  1 Logarithms 5 (10) log p a  log p b  a  b if base p is positive and >1 or a  b if base p is positive and < 1 i.e., 0  p 1 1 then the sign of inequality is reversed. If x  log 3 5, y  log 17 25, which one of the following is correct Solution: (c) x y (b) (c) x y y  log 17 25  2 log 17 5  1 1  log 5 17 y 2 1 1  log 5 3  log 5 9 x 2 1 1 ,x y  y x U Clearly x y (d) None of these E3 (a) [W.B. JEE 1993] ID Example: 8 60 In other words, if base is greater than 1 then inequality remains same and if base is positive but less than If log 0.3 (x  1)  log 0.09 (x  1), then x lies in the interval (a) Solution: (a) ( 2,  ) (b) (– 2, –1) log 0.3 (x  1)  log (0.3 )2 (x  1)   (c) (1, 2) D YG Example: 9 (d) None of these 1 log 0.3 (x  1) 2 1 log 0.3 (x  1)  0 2 or log 0.3 (x  1)  0  log 1 or (x  1)  1 or x  2 ST U As base is less than 1, therefore the inequality is reversed, now x>2  x lies in (2, ). ***