**Logarithms**

In general Terms, **Logarithm** refers to the number that shows how many times a base number (such as ten) is multiplied by itself to produce a third number (such as 100).

Logarithm was developed by John Napier and is very useful tools for mathematical computation.

If n^{x}= a, the *logarithm* of a, with n as the base, is x. It is symbolically expressed as *log*_{n} a = x.

The equations, a^{x}=n and x=log_{a}n are only transformation of each other, explaining the relationship between the 3 components, base, index & logarithm.

Logarithm may be defined as the number, which, to a given base as the index (or the power) to which the Base should be raised to yield the given number.

In the expression, 2^{4} = 16, 2 is termed to as the base, 4 as the index (or power), 16 is the number. This could also be expressed as log_{2}16 = 4 and is read as the Logarithm of 16 to base 2 is 4.

In the expression 5^{-3} = 1/5^{3 }=1/125, So, log_{5} (1/125) = log _{5}(1/5^{3}) = log_{5}(5^{-3}). So, log_{5} (5^{-3}) = -3

**Logarithm Rules**

- The logarithm of 1 on any base is 0.Ex. Log
_{10}1=0, Log_{2}1=0 - The logarithm of a number with same base is 1. Ex. log
_{13}13 = 1. - The logarithm of a product to base a is equal to sum of the logarithms of the individual numbers which constitute the product to the same base a.

log_{n}(AxB) = log_{n}A + log_{n}B. Ex. Log_{n} (15) = Log _{n} (5X3) = Log _{n} 5 + Log_{n} 3

- The logarithm of a fraction to the base a will be equal to the difference of the logarithm

of the numerator to the base a, and the logarithm of the denominator to base n.

log_{n }(A/B) = log_{n}A − log_{n}B. Ex. Log _{n} 25 = Log _{n} (100/4) = Log _{n} 100 – Log _{n} 4

- The logarithm of a number raised to any power, integral or fractional, is equal to product of that number and the logarithm of the number raised to base a.

log_{n }(M^{p}) = p.log_{n }M. Ex Log_{n} 5^{2 }= 2 Log_{n} 5

**Natural Base** : 10 is considered as natural base (common logarithm). If no Base is mentioned, it is assumed that the Base is 10. So, log 234 indicates log _{10} 234

**Logarithm Rule – Problem**

**Rules of Logarithm – Problems & Solutions**

**Ex.** Solve 7 log (32/15) + 5 log (25/24) + 3 log (81/80).

Step 1= 7 (log 32 – log 15) + 5 (log 25 – log 24) + 3 (log 81 – log 80)

Step 2= [7 {log 2^{5} – log (3 x 5)}] + [5 {log 5^{2} – log (2^{3}x3)}] + [3 {log 3^{4} –log (2^{4}x5)}]

Step 3= {7{5 log 2– (log 3 + log 5)}] + [5 {2log 5 – (3log 2 +log 3)}] + [3{4log 3 – (4 log 2 + log 5)}]

Step 4= [7 (5 log 2 – log 3 – log 5)] + [5 (2 log 5 – 3 log 2 – log 3)] + [3(4 log 3 – 4 log 2 – log 5)]

Step 5= 35 log 2 – 7 log 3 – 7 log 5 + 10 log 5 – 15 log 2 – 5 log 3 + 12 log 3 – 12 log 2 – 3 log 5

Step 6= 35 log 2 – 15 log 2 – 12 log 2 – 7 log 3 – 5 log 3 + 12 log 3 – 7 log 5 + 10 log 5 – 3 log5

Step 7= 35 log 2 – 27 log 2 – 12 log 3 + 12 log 3 – 10 log 5 + 10 log 5

Step 8= (35 – 27) log 2 + (−12 + 12) log 3 + (− 10 + 10) log 5

Step 9= 8 log 2 + 0 log 3 + 0 log 5

Step 10= 8 log 2 + 0 + 0

Step 11= 8 log 2.

**Logarithms Rule – Problems**

= log_{2} (25 / 100)=log_{2} (1/ 4) = log_{2} (1/ 2^{2})= log_{2} (2^{-2}) = -2

log [(abc^{2}) / d^{3}] = (log a + log b + log c^{2}) – log (d^{3}) = log a + log b + 2log c – 3log d

log[(x√5)/z^{5}] = logx + log (√5) – log (z^{5}) = log x + log (5^{1/2 }) – 5 log z= log x +½(log5 )- 5 log z

**Change of Logarithm Base**

Often, you need to **change logarithm base** to get the logarithm of a number from one Base to another base.

The relation between the logarithms of a number to different base

Log_{a} m = log _{b}m x log _{a}b

logarithm of a number (m), with existing base(a), i.e Log_{a} m = Logarithm of the number (m) with new Base (b) x Logarithm of the new base (b), with respect to the existing base (a)

**Ex.** Change the base log_{2} 45 to common logarithmic base.

Log_{a}m = log_{b}m / Log_{b}a, So, log_{2}45 = Log_{10}45 / Log_{10}2 = log_{10}45 – log_{10}2

**Properties of Logarithm**

- log
_{a}mn =log_{a}m + log_{a}n. Logarithm of product of two number is equal to sum of the logarithm of each number. - log
_{a }(m/n) =log_{a}m – log_{a}n. Logarithm of fraction of two number is equal to difference between the Logarithm of numerator and Logarithm of Denominator - log
_{a}m^{n}=n(log_{a}m) : Logarithm of power (exponential) of number is equal to product of the power (exponential) and the Logarithm of the number - log
_{a}a = 1 : Logarithm of a number with same base is always 1 - log
_{a}1 = 0 : Logarithm of 1 is always zero (irrespective of value of Base), because a^{0}=1, for any value of a. - log
_{a}m = 1/ (log_{m}a). So, log_{a}m X log_{m}a = 1. When Base and the Number is interchanged, the logarithm value is inversed, and the value of their product is always 1 - log
_{b}n = ( log_{a}n) / (log_{a}b) : When Base is changed, the logarithm value of the number with new base is the ratio between the logarithm value of the number with current base and the logarithm value of the Base with Base of the number - log
_{a}b x log_{b}c x log_{c}a =1 : The product of logarithm of 3 numbers in series with their Base is 1 - log
_{a}m^{n }= (n/b) log_{b}m : The logarithm of a number raised to a power (exponent) is equal to the product of the number divided by the new Base and logarithm of the number with new Base

**Properties of Logarithm – Problems**

**Properties of Logarithm – Problems & Solutions**

**Ex.** Find the value of log_{2}32 – log_{8}2

Step 1 : log_{2}(2^{ 5 }) – [1/ (log_{2}8)]

Step 2 : log_{2} 2^{5} – [1/ log_{2} (2^{3})]

Step 3 : 5 (log_{2} 2) – {1/ (3 log_{2} 2) = (5×1) -1/(3×1)= 5-1/3= 14/3

So, the value of the expression is 14/3

**Logarithm & Antilogarithm Tables**

**Logarithm Table** ( Log Table) & Antilogarithm Tables (Anti Log Table) are were wisely used in performing large calculations (of multiplication, division, squares, and roots) without using a calculator.

Log Table & Antilog Tables are printed in Paper, from which the value of Logarithm and then the reverse Antilogarithm is found. However, with advent of electronic calculator, now more complex arithmetic an scientific calculations can be performed using such digital calculators. So, Log Tbales are outdated and no more used.

However, a brief commentary and Illustration on Log Table & Antilog Tables are included to comply syllabus still included in some courses.

**Logarithm & Antilogarithm Tables** **– Problems**

**Logarithm & Antilogarithm Tables** **– Problems and Solutions**

**Ex.**Find the value of ∜0.06754.

Let x =∜0.06754= (0.06754)^{1/4}

Taking log on both sides, we get

log x = (1/4) x log(0.06754)

log x = (1/4) x (–2 + .8296)

log x = (1/4) x {–2 + (-2+2)+ .8296}

log x = (1/4) x [{–2 + (-2)}+ (2)+ .8296)}

log x = (1/4) x (-4+ 2.8296)

log x = [{(1/4) x (-4)} + {(1/4)x2.8296)}]

log x = -1 + .7074

So, x = antilog (-1 + .7074) = 0.5098

**Properties of Logarithm – Problems**

**Logarithm – Problems and Solutions**

**Ex.** If log_{a}bc = x, log_{b}ca = y, log_{c}ab = z, prove that

{(1/(x+1)} + {(1/(y+1)} + {(1/(z+1)} + 4=5

Step 1 : We may write

x+1 = log_{a}bc + 1 = log_{a}bc + log_{a}a = log_{a}abc

y+1 = log_{b}ca + 1 = log_{b}ca + log_{b}b = log_{b}abc

z+1 = log_{c}ab + 1 = log_{c}ab + log_{c}c = log_{c}abc

Step 2 : So, the LHS of the above expression may be written as

LHS= {(1/ log_{a}abc) + (1/ log_{b}abc) + (1/ log_{c}abc)} +4

LHS = (log_{abc}a + log_{abc}b + log_{abc}c) +4

LHS = log_{abc}abc +4 = 1+ 4 = 5

**Characteristics and Mantissa in Log Tables**

**Common logarithms : **Logarithms calculated to the base 10 (Natural Base) are called common logarithms. Common logarithm has two parts, the whole part and the decimal part.

**Characteristic**: The whole part of the logarithm of a number is called Characteristic**Mantissa.:**The non-negative decimal part is called the Mantissa.

**Example** : When value of logarithm is 2.4567. The value of Characteristic is 2 and value of Mantissa is .4567

**Determination of Characteristics**

**Characteristics****of Number greater than 1**: When the logarithm of a number is greater than 1, first determine number of digit before decimal point. Then deduct 1 from number of digits before decimal point to get the get the characteristics.

**Examples : **Characteristics of 2.454. 2.454> 1. number of digit before the decimal point = 1**. **So**, **characteristic of log 2.454= 1 – 1 =0

Characteristics of 245.8. 245.8> 1. The number of digit before the decimal point = 3. Hence, characteristic of log 245.8= 3 – 1 =2

**Characteristics****of Number less than 1**: Logarithm of a number less than 1 is negative and the characteristic value is one more than the number of zeroes immediately before the decimal point.

**Examples** : Characteristics of .2454 The number .2454<1. Number of zeroes before the decimal point = 0. So, characteristic of log .2454= – (0 +1) = -1 (or`1 )

Characteristics of.002454 . The number .002454 <1. The number of zeroes before the decimal point = 2. So, characteristic of log .002454 = – (2 +1) = -3 (or `3 )

**Computation of Mantissa**

The Mantissa is same for the same figures in the same order and does not depend on the position of the decimal point.

Only significant figure is considered and the Mantissa of the logarithm of a number is determined by using logarithm table.

**Ex.** Mantissa of 2454, 2.454, .2454, .0002454, is same and its value can be determined from the log table.

**Finding value of Mantissa from Log Table.**

The main body of the Logarithmic Table (referred as Log Table) gives the mantissa of the logarithm of numbers of 3 digits or less, while the mean difference Table provides the increment for the fourth digit.

**Step 1 : **Ignore the decimal point (if any) of the given number.

**Step 2 : **In the first column (from left) in the table, locate the first two digits of the given number. If however, the given number consists only one digit, we make it 2 digited number by adding a zero its right.

**Step 3 :** Along the row corresponding to these two digits, locate the entry under the column headed by the third digit in the given number.

**Step 4 : **For the fourth digit, move further along that row and suitably locate the entry in the column (of mean difference) header of the fourth digit. Now, add this entry to the number obtained in step 3.

**Step 5 : **The final entry is thus obtained from the table, preceded by a decimal point is the required mantissa of the given number.

**Determination of Logarithm – Problems**

**Computation of Logarithm using Log Tables**

Find the value of log 24.57.

**Determination of characteristics : **24.57>1. Number of digits before decimal place is 2. So, the Characteristics is 2-1 = 1

**Determination of Mantissa from Log Table**

**Step 1 : **Ignore the decimal point first.

**Step 2 : **Locate24(the first 2 digit ) in the first column.

**Step 3 : **Move along the row of 24 and locate the entry in the column head 5 (third digit). This entry is 3892.

**Step 4 : **Move further along the same row and locate the entry in column of Mean Difference under column head 7 (the fourth digit). This entry is 12.

**Step 5 : **Add 3892 and 12 to get the value of Mantissa, i.e 3904.

**Step 6: **Write the characteristics (i.e 1). Put decimal point and write the mantissa. So the value of log 24.57 = 1.3904.

**Antilogarithms**

If x is the logarithm of a given number n with a given base, then n is called the **antilogarithm** (antilog) of x to that base.

If log(n) = x, then n = antilog x

If log 2457 = 3.3904 then Antilog 3.3904 = 2457

**Ex.** Find the value of log 50 if log 2 = .3010

Log 50 = log (100/2) = Log 100 – log 2 = 2-.3010 = 1.6990

**Determination of Antilog value from Antilog Table**

**– Step 1** : Ignore the characteristic of the number (Logarithmic Value) for which Antilog is to computed. Make sure that decimal part is positive.

**– Step 2 : **Locate the value in the antilog table for the value (the process is similar to finding from Log Table).** **

**– Step 3** : Insert the decimal point in the value obtained from Anti Log Table. The number of digits before decimal point will be 1 more than the characteristic of Logarithmic value (the number for which the antilog value is being computed)

**Ex.** Find Antilog 1.3246.

From the antilog table, for mantissa .324, the number = 2109. For mean difference 6, the number = 3. So, for mantissa .3246, the number = 2112

The characteristic is 1, therefore, the number must have 2 digits in the integral part.

Hence, Antilog 1.3246 = 21.12

**Ex.** Find Antilog (– 2.3246).

−2.3246 = −3 + (3 − 2.3246) = (−3) + .6754. Now find Antilog for .6754. From Antilog Table, you get 7332 for .675 and then the mean difference for 4 is 4. So, adding together, you get Antilog (.6754) = 7332+4=7336. The characteristic is (-3).

So, the number is less than 1 and there must be 2 zeros before the decimal place. So, the value would be 0.007336. So, Antilog (−2.3246) = 0.007336

**Computation of Logarithm and Antilogarithm**

**Computation using Logarithm and Antilogarithm Tables**

**Ex.** **Find the value of **(391.1 x (0.08354)^{1/5} / (3.843)^{3}

y = [{**(**391.1 x (0.08354)^{ 1/5}} / (3.843)^{3}]

Taking log on both sides, we get

Log y = log 391.1 + [(1/5) x (log 0.8354)] – {3 (log 3.843)}

LHS = 2.5923 + [(1/5) {(-1 + 0.9219)}] – [3 x (.5847)]

LHS = 2.5923 + [(1/5) {(-5 + 4.9219)] – 1.7541

LHS = 2.5923 + [(-1 + 0.09844) – 1.7541]

LHS = 2.5923 + (-1) + 0.09844 – 1.7541

LHS = 2.5923 -1 + 0.09844 – 1.7541 = 0.8226

So, y= antilog (0.8226) = 6.646

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