Chapter-23 (LOGARITHMS)



23. LOGARITHMS

IMPORTANT FACTS AND FORMULAE


I.                   Logarithm: If a is a positive real number, other than 1 and am = X, then we write:
          m = loga x and we say that the value of log x to the base a is m.
Example:
(i) 103 = 1000 => log10 1000 = 3
(ii) 2-3 = 1/8 => log2 1/8 = - 3
(iii) 34 = 81 => log3 81=4
(iiii) (.1)2 = .01 => log(.l) .01 = 2.

II. Properties of Logarithms:

1. loga(xy) = loga x + loga y
2. loga (x/y) = loga x - loga y
3.logx x=1
4. loga 1 = 0
5.loga(xp)=p(logax)       1
6. logax =­1/logx a
7. logax = logb x/logb a=log x/log a.
                          
Remember: When base is not mentioned, it is taken as 10.

II.                Common Logarithms:
         Logarithms to the base 10 are known as common logarithms.

III.             The logarithm of a number contains two parts, namely characteristic and mantissa.
Characteristic: The integral part of the logarithm of a number is called its characteristic.

Case I: When the number is greater than 1.
In this case, the characteristic is one less than the number of digits in the left of the decimal point in the given number.

Case II: When the number is less than 1.
In this case, the characteristic is one more than the number of zeros between the decimal point and the first significant digit of the number and it is negative.
Instead of - 1, - 2, etc. we write, `1 (one bar), `2 (two bar), etc.



Example:

Number
Characteristic
Number
Characteristic
348.25
2
0.6173
`1
46.583
1
0.03125
`2
9.2193
0
0.00125
`3

Mantissa: The decimal part of the logarithm of a number is known is its mantissa. For mantissa, we look through log table.




SOLVED EXAMPLES


1.Evaluate:
(1)log3 27
(2)log7 (1/343)
(3)log100(0.01)
SOLUTION:
(1)   let log3 27=3­­­­3 or n=3.

ie, log3 27 = 3.

(2)   Let log7 (1\343) = n.

Then ,7n ­=1/343
              =1/73
                    n = -3.
    ie,
     log7(1\343)= -3.
(3)   let log100(0.01) = n.
     

Then,. (100) = 0.01 = 1 /100=100 -1 0r n=-1

EX.2. evaluate

(i) log7 1=0            (ii)log34 34     (iii)36log 6 4
solution:
i)                    we know that loga 1=0 ,so log7 1=0 .
ii)                  we know that     loga a=1,so log34 34=0.
      iii)       We know that     alog 6 x =x.
                  now  36log 6 4=(62)log6   4 =6 log 6(16)=16.



Ex.3.if log  x=3 (1/3), find the value of x.
        
 log  x=10/3 ,x=()10/3=(23/2)10/3=2(3/2*10/3)=25=32.

Ex.4:Evaluate: (i) log53*log27 25 (ii) log 27 –log27 9

(i)log 53 * log27 25=(log 3/log 5)*(log 25/log 27)
                            =(log 3/log 5)*(log 52*log33)
                            =(log 3/log 5)*(2log5/3log3)
                            =2/3


(ii)Let log927=n
Then,
9n =27   ó32n  =3 3    ó2n=3ó n=3/2
Again, let log279=m
Then,
27m =9   ó33m  =3 2    ó3m=2ó m=2/3

ð  log927- log279=(n-m)=(3/2-2/3)=5/6


Ex 5. Simplify :(log 75/16-2 log 5/9+log 32/243)
Sol: log 75/16-2 log 5/9+log 32/243
= log 75/16-log(5/9)2+log32/243
= log 75/16-log25/81+log 32/243
= log(75/16*32/243*81/25)=log 2


Ex. 6.Find the value of x which satisfies the relation

Log10 3+log10 (4x+1)=log10 (x+1)+1
Sol: log10 3+log10 (4x+1)=log10 (x+1)+1
Log10 3+log10 (4x+1)=log10 (x+1)+log10 (x+1)+log10 10
Log10 (3(4x+1))=log10 (10(x+1))
=3(4x+1)=10(x+1)=12x+3
=10x+10
=2x=7=x=7/2

Ex. 7.Simplify:[1/logxy(xyz)+1/logyz­(xyz)+1/logzx(xyz)]
Given expression: logxyz xy+ logxyz yz+ logxyz zx
=logxyz (xy*yz*zx)=logxyz (xyz)2
  2logxyz(xyz)=2*1=2

Ex.8.If log10 2=0.30103,find the value of log10 50.
  Soln. log10 50=log10 (100/2)=log10 100-log10 2=2-0.30103=1.69897.

Ex 9.If log 2=0.3010 and log 3=0.4771,find the values of:
i)                    log 25       ii)log 4.5
  Soln.
i)                    log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398.
ii)                  log 4.5=log(9/2)=log 9-log 2=2log 3-log 2
                        =(2*0.4771-.3010)=.6532

Ex.10. If log 2=.30103,find the number of digits in 256.
    Soln.     log 256  =56log2=(56*0.30103)=16.85768.
 Its characteristics is 16.
Hence,the number of digits in 256 is 17


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