LOGARITHMS 

IMPORTANT FACTS AND FORMULAE

1. Logarithm: If a is a positive real number, other than 1 and a^m = X, then we write:

          m = log_a 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. log_a(xy) = log_a x + log_a y
2. log_a (x/y) = log_a x – log_a y

3.log_x x=1

4. log_a 1 = 0

5.log_a(x^p)=p(log_ax)       1

6. log_ax =­1/log_x a
7. log_ax = log_b x/log_b a=log x/log a.

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

1. Common Logarithms:

         Logarithms to the base 10 are known as common logarithms.

• The logarithm of a number contains two parts, namely characteristic and

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)log_100(0.01)

_SOLUTION:

• 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.

• 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)36^log ₆ 4

solution:

1. we know that log_a 1=0 ,so log₇ 1=0 .
2. we know that log_a a=1,so log₃₄ 34=0.

      iii)       We know that     a^log ₆ x =x.

                  now  36^log 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=(2^3/2)^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,

27^m =9   ó3^3m  =3 ²    ó3m=2ó m=2/3

• 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/log_xy(xyz)+1/log_yz­(xyz)+1/log_zx(xyz)]

Given expression: log_xyz xy+ log_xyz yz+ log_xyz zx

=log_xyz (xy*yz*zx)=log_xyz (xyz)²

  2log_xyz(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. log 25 ii)log 4.5

  Soln.

1. log 25=log(100/4)=log 100-log 4=2-2log 2=(2-2*.3010)=1.398.
2. 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

“Quantitative Aptitude for Competitive Examinations” by R.S. Aggarwal is a comprehensive resource widely used by aspirants preparing for various competitive exams. The book includes a dedicated chapter on Logarithms, providing detailed explanations, solved examples, and practice questions to enhance understanding of the topic.

Key Features of the Logarithms Chapter:

• Conceptual Clarity: The chapter begins with fundamental definitions and properties of logarithms, ensuring a solid foundation.

• Solved Examples: Numerous examples are provided to illustrate the application of logarithmic concepts in problem-solving.

• Practice Questions: A variety of questions are included to test comprehension and improve problem-solving speed and accuracy.

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