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23. LOGARITHMS
IMPORTANT FACTS AND FORMULAE

(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

nl

5.loga(xp)=p(logax) 1
6. logax =1/logx a
7. logax = logb x/logb a=log x/log a.

eG
K

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

in

I.

Remember: When base is not mentioned, it is taken as 10.

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

eO

II.

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.

Th

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:
Characteristic
2
1
0

Number
0.6173
0.03125
0.00125

Characteristic
1
2
3

eG
K

Number
348.25
46.583
9.2193

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

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

(1) let log3 27=33 or n=3.

nl

ie, log3 27 = 3.

in

SOLVED EXAMPLES

eO

(2) Let log7 (1343) = n.
Then ,7n =1/343
=1/73
n = -3.
ie,
log7(1343)= -3.
(3) let log100(0.01) = n.

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

Th

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.

(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 32n =3 3 2n=3 n=3/2
Again, let log279=m
Then,
27m =9 33m =3 2 3m=2 m=2/3

in

 log927- log279=(n-m)=(3/2-2/3)=5/6

eG
K

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

nl

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

Th

eO

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)
Soln.
i)
ii)

log 25

ii)log 4.5

log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398.
log 4.5=log(9/2)=log 9-log 2=2log 3-log 2
=(2*0.4771-.3010)=.6532

Th

eO

nl

in

eG
K

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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23. logarithms

  • 1. Free GK Alerts- JOIN OnlineGK to 9870807070 23. LOGARITHMS IMPORTANT FACTS AND FORMULAE (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 nl 5.loga(xp)=p(logax) 1 6. logax =1/logx a 7. logax = logb x/logb a=log x/log a. eG K 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 in I. Remember: When base is not mentioned, it is taken as 10. Common Logarithms: Logarithms to the base 10 are known as common logarithms. eO II. 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. Th 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.
  • 2. Example: Characteristic 2 1 0 Number 0.6173 0.03125 0.00125 Characteristic 1 2 3 eG K Number 348.25 46.583 9.2193 Mantissa: The decimal part of the logarithm of a number is known is its mantissa. For mantissa, we look through log table. 1.Evaluate: (1)log3 27 (2)log7 (1/343) (3)log100(0.01) SOLUTION: (1) let log3 27=33 or n=3. nl ie, log3 27 = 3. in SOLVED EXAMPLES eO (2) Let log7 (1343) = n. Then ,7n =1/343 =1/73 n = -3. ie, log7(1343)= -3. (3) let log100(0.01) = n. Then,. (100) = 0.01 = 1 /100=100 -1 0r n=-1 Th 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.
  • 3. log x=10/3 ,x=( )10/3=(23/2)10/3=2(3/2*10/3)=25=32. (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 32n =3 3 2n=3 n=3/2 Again, let log279=m Then, 27m =9 33m =3 2 3m=2 m=2/3 in  log927- log279=(n-m)=(3/2-2/3)=5/6 eG K Ex.4:Evaluate: (i) log53*log27 25 (ii) log 27 –log27 9 nl 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 Th eO 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:
  • 4. i) Soln. i) ii) log 25 ii)log 4.5 log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398. log 4.5=log(9/2)=log 9-log 2=2log 3-log 2 =(2*0.4771-.3010)=.6532 Th eO nl in eG K 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