log

Contents

1. INTRODUCTION
2. COMMON LOGARITHMS
3. NATURAL LOGARITHMS
4. LAWS AND PROPERTIES OF LOGARITHMS
5. THE CHARACTERISTIC AND THE MANTISSA
6. SOLVED EXAMPLES

Logarithms

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INTRODUCTION

Long before the invention of calculators or any other computing device, mathematicians used logarithms to make complex calculations.

In words, p is referred to as the logarithm of N to the base a. Here, a cannot be negative, 0 or 1 because logarithm of negative numbers and 0 is not defined and the logarithm of 1 (to the base 10) is equal to 0 ( a⁰ = 1).

In other words, the logarithm of any number to a given base is the power to which the base must be raised to get that number.

The reverse is also true. That is, if log_a

N = p, then we can write N = a^p

For example,

Since 81 = 3⁴,

log₃ 81 = 4

REMEMBER:

• Logarithms are defined only for positive numbers. It is not defined for zero or negative numbers.

• Logarithms can be expressed in any base.

• Logarithms expressed in one base can be converted to logarithms expressed in any other base.

• Logarithms are generally expressed to the base 10. These are called common logarithms.

We will now study them in detail.

Example 1:

Then what is the value of x ? [FMS 2009] (1) 2 (2) 3 (3) 4 (4) 5 Solution: If log₇ a = 0, then a = 7⁰. Thus a = 1. On substituting the values of x from the given options, we see that x = 4 satisfies the above equation. Hence, option 3.

COMMON LOGARITHMS

Common logarithms are expressed in the base 10. Most problems in logarithms deal with common logarithms. If no base is mentioned, it is assumed that the base is 10.

For example, the common logarithm of 100 would be expressed as log₁₀ 100 and is equal to 2.

( 100 = 10²)

The following table gives the values of common logarithms for some select values.

N	log10 N
0.001	–3
0.01	–2
0.1	–1
1	0
10	1
100	2
1000	3
10000	4

Shown below is a graph of the function y = log₁₀ x

NATURAL LOGARITHMS

Logarithms are also expressed in the base ‘e’. Such logarithms are called Natural logarithms or Napierian logarithms.

e is an irrational constant approximately equal to 2.718281828.

The natural logarithm of a positive real number x, is represented by ln(x) or log_e(x).

Like common logarithm, natural logarithm is also not defined for zero and negative real numbers .

LAWS AND PROPERTIES OF LOGARITHMS

Most problems involving logarithms can be solved by applying the following laws/properties:

1. The logarithm of 1 to any base is always 0.

i.e. log_b

1 = 0 (provided that b is positive and ≠ 1)

Explanation:

For any real number b, b⁰ is defined as 1.

Hence, log_b

1 = 0

2. The logarithm of a number to a base which is equal to that number is 1.

i.e. log_b b = 1

(provided that b is positive and ≠ 1)

Explanation:

For any real number b, b¹

= b

Hence, log_b b = 1

Example 2:

[FMS 2010] (1) 25(2) 10 (3) 2 (4) 1 Solution: = 12 = 1 Hence, option 4.

3. log_a b log_b a = 1

Explanation:

Let log_a b = x

a^x = b

a = b¹/x

log_b a = 1/x = 1/(log_a b)

log_a b log_b a = 1

4. The logarithm of the product of two numbers is equal to the sum of the logarithms of the individual numbers.

i.e. log_b

(m n) = log_b m + log_b n

Example 3:

[CAT 2002] Solution: Hence, option 4.

Example 4:

What is the sum of n terms in the series: [CAT 2003 Re-test] Solution: Hence, option 4.

5. The logarithm of the ratio of numbers is equal to the difference obtained when the logarithm of the denominator is subtracted from that of the numerator.

Explanation:

Let log_b

m = x and log_b n = y

m = b^x

and n = b^y

mn = b^x

b^y = b^x + y

log_b

(mn) = x + y = log_b m + log_b n

Similarly, we can prove that,

Example 5:

Find the value of log9 171 – log9 19? Solution: … using law 2

6. The logarithm of a number raised to a power is equal to the product of the power and the logarithm of the number.

i.e. log_b

(mⁿ) = n log_b m

Explanation:

Let log_a

mⁿ = log_a (m m m … m)

Using the law 4 above, the RHS can be expressed as,

log_a

(m m m … m) = log_a m + log_a m + log_a m + … + log_a m = n log_a m

Hence, log_a

mⁿ = n log_a m

7. Logarithms expressed in one base can be converted to logarithms expressed in any other base.

Explanation:

Let log_b

m = x

m = b^x

Let a^y

= b

y = log_a

b

If we substitute the value of b in the expression for m, we get,

m = (a^y

)^x = a^xy

log_a

m = xy = log_b m log_a b

log_b

m = log_a m/log_a b

= log_a

m log_b a ...(from law 3)

8. 

Explanation:

Let log_b

n = x

n = b^x

Substituting the value of x in the above expression, we get,

Example 6:

Solution:

9. Given an equation, log_a

M = log_b N:

1. If M = N, then a will be equal to b.

2. If a = b, then M will be equal to N.

10. In order to compare two numbers, when the comparison between their respective logarithms is given:

1. If b > 1 and log_b

m > log_b n, then m > n

2. Also, if b < 1 and log_b

m > log_b n, then m < n.

THE CHARACTERISTIC AND THE MANTISSA

The logarithm of a number has an integral part and a decimal part. The integral part is called the characteristic and the decimal part is called the mantissa.

For example, log₁₀ 9100

= log₁₀ (10³ 9.1)

log₁₀ 10³ + log₁₀ 9.1

= 3 + 0.959

Here, 3 is the characteristic and 0.959 is the mantissa.

An important property of the characteristic (for common logarithms) is that (characteristic + 1) will give the number of digits of the integral part of the number whose logarithm you determined. In the above example, 9100 contains (3 + 1) = 4 digits.

This is explained as follows:

log₁₀ 9100 = 3.959,

9100 = 10^3.959 = 10^{3 + 0.959} = 10³ 10^0.959

= 10^x

10^y

Since y will always be < 1, 10^y

will always be < 10. Hence, the number of digits in the number will be equal to the number of digits in 10^x, which in turn is equal to (x + 1).

Also, the characteristic of a logarithm could be either positive or negative; however, the mantissa should always be positive.

For example, log₁₀ 0.091 = log₁₀ (10^–2 9.1)

–2 + 0.959 =

If the calculated value of the logarithm of some number is negative (usually, calculators generate values containing a negative mantissa), then we make the mantissa positive as follows:

–1.041 = –(1 + 0.041) = –1 – 0.041

= (–1 – 1) + (–0.041 + 1)

= –2 + 0.959

=

Example 7:

If the logarithm of a number is 3.153, what are Characteristic and Mantissa? [FMS 2009] (1) Characteristic = 4, mantissa = 0.847 (2) Characteristic = 3, mantissa = 0.153 (3) Characteristic = 4, mantissa = 0.847 (4) Characteristic = 3, mantissa = 0.153 Solution: The characteristic can either be positive or negative; however, the mantissa always remains positive. In the given answer options, only option 1 has a positive mantissa. Here, 3.153 = (3 + 0.153) = 3 0.153 = (3 1) + (1 0.153) = 4 + 0.847 Hence, option 1.

Example 8:

Find the number of digits in 162521, given that log 2 = 0.301 and log 1.3 = 0.114 Solution: log (162521) = 21 log 1625 = 21 log (125 13) = 21 (log 125 + log 13) = 21 (log 53 + log 13) = 21 (3 log 5 + log 13) = 21 [3 log (10/2) + log (10 1.3)] = 21 [3 (log 10 – log 2) + log 10 + log 1.3] = 21 [3 (1 – log 2) + 1 + log 1.3] = 21 [(3 (1 – 0.301) + 1 + 0.114] = 21 3.211 = 67.431 Therefore, we can say that 1067 < 162521 < 1068 Hence, the number of digits in the number will be 68.

SOLVED EXAMPLES

Example 9:

Simplify log3 9 + log3 3 Solution: log3 9 + log3 3 = log3 27… using law 4 = log3 33 =

3log₃ 3… using law 6
= 3… using law 2

Example 10:

Find the value of (log 3000 + log 400 – 3 log 25), if log 3 = 0.4771, log 2 = 0.301 and log 5 = 0.699. Solution: The base of the logarithms is 10. log 3000 + log 400 – 3 log 25 = log (3 103) + log (22 102) – 3 log 52 =

log 3 + log 10³ + log 2² + log 10² – 6 log 5

= log 3 + 3 log 10 + 2 log 2 + 2 log 10 – 6 log 5

= log 3 + 3 + 2 log 2 + 2 – 6 log 5

= log 3 + 2 log 2 – 6 log 5 + 5

= 0.4771 + 2 0.301 – 6 0.699 + 5

= 1.8851

Example 11:

Solve for x: log [(x – 1)(x + 1)] = log 8 Solution: log [(x – 1)(x + 1)] = log 8 (x + 1)(x – 1) = 8 x2 – 1 = 8 x2 = 9 x = +3 or x = –3

Example 12:

Solution: = 3

Example 13:

If a, b, c are in G.P. and = = then: [FMS 2009] Solution: In this problem, the answer option does not contain a, b or c. Hence we will need to eliminate a, b and c and get the answers in terms of x, y and z. This is done easily using logarithms. a, b and c are in G.P b² = a c Taking log on both sides, we get, 2 log b = log a + log c … (i) Taking log throughout, we get, x log a = y log b = z log c To eliminate a, b and c, we should express any 2 of a, b or c in terms of the 3rd variable. Here, we will express a and c in terms of b. Substituting the same in equation (i), we get, Hence, option 1.