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## Logarithms

Let a be a positive real number, a ≠ 1 and a

^{x}= m. Then x is called the logarithm of m to the base a and is written as log_{a}m, and conversely, if log_{a}m = x, then a^{x}= m.

Note: Logarithm to a negative base is not defined.

Also, logarithm of a negative number is not defined. Hence, in the above logarithmic equation, log

_{a}m = x, and we can say that m > 0 and a > 0.

Thus a

^{x}= m so x = log_{a}m and log_{a}m = x so a^{x}= m

In short, a

^{x}= m so x = log_{a}m.

x = log

_{a}m is called the logarithmic form and a^{x}= m is called the exponential form of the equation connecting a, x and m.

**Two Properties of Logarithms :**

1. log

_{a}1 = 0 for all a > 0, a ≠ 1 That is, log 1 to any base is zero

2. log

_{a}a = 1 for all a > 0, a ≠ 1 That is, log of a number to the same base is 1

**Laws of Logarithms :**

**First Law:**log_{a}(mn) = log_{a}m + log_{a}n That is, log of product = sum of logs

**Second Law:**log_{a}(m/n) = log_{a}m – log_{a}n That is, log of quotient = difference of logs

Note: The first theorem converts a problem of multiplication into a problem of addition and the second theorem converts a problem of division into a problem of subtraction, which are far easier to perform than multiplication or division. That is why logarithms are so useful in all numerical calculations.

**Third Law:**log_{a}m^{n}= n log_{a}m

**Generalisation**

**log (mnp) = log m + log n + log p**

**log(a1a2a3...ak)=loga1+loga2+...+logak**

Note: Common logarithms: We shall assume that the base a = 10 whenever it is not indicated. Therefore, we shall denote log

_{10}m by log m only. The logarithm calculated to base 10 are called common logarithms.

**The Characteristic and Mantissa of a Logarithm**

The logarithm of a number consists of two parts: the integral part and the decimal part. The integral part is known as the characteristic and the decimal part is called the mantissa.

For example, In log 3257 = 3.5128, the integral part is 3 and the decimal part is .5128; therefore, characteristic = 3 and mantissa = .5128.

It should be remembered that the mantissa is always written as positive.

Rule: To make the mantissa positive (in case the value of the logarithm of a number is negative), subtract 1 from the integral part and add 1 to the decimal part.

**Thus,–3.4328 = – (3 + .4328) = –3 – 0.4328 = (–3 –1) + (1 – 0.4328) = –4 + .5672. so the mantissa is = .5672.**

Note: The characteristic may be positive or negative. When the characteristic is negative, it is represented by putting a bar on the number.

Thus instead of –4, we write . Hence we may write –4 + .5672 as .5672.

**Base Change Rule**

Till now all rules and theorems you have studied in Logarithms have been related to operations on logs with the same basis. However, there are a lot of situations in Logarithm problems where you have to operate on logs having different basis. The base change rule is used in such situations. This rule states that

**(i) log**_{a}(b) = log_{c}(b)/log_{c}(a)

**It is one of the most important rules for solving logarithms. (ii) log**_{b}(a) = log_{c}(a)/log_{b}(c)**A corollary of this rule is (iii) log**_{a}(b) = 1/log_{b}(a)

(iv) log c to the base a

^{b}is equal to logacb.

**Results on Logarithmic Inequalities :**

(a) If a > 1 and log

_{a}x_{1}> log_{a}x_{2}then x_{1}> x_{2}

(b) If a < 1 and log

_{a}x_{1}> log_{a}x_{2}then x_{1}< x_{2}

**Applied conclusions for logarithms**

The characteristic of common logarithms of any positive number less than 1 is negative.

The characteristic of common logarithm of any number greater than 1 is positive.

**If the logarithm to any base a gives the characteristic n, then we can say that the number of integers possible is given by a**.^{n + 1}– a^{n}

Example: log

_{10}x = 2.bcde…, then the number of integral values that x can take is given by: 10^{2 + 1}– 10^{2}= 900. This can be physically verified as follows.

**Log to the base 10 gives a characteristic of 2 for all 3 digit numbers with the lowest being 100 and the highest being 999. Hence, there are 900 integral values possible for x.**

**If –n is the characteristic of log**_{10}y, then the number of zeros between the decimal and the first significant number after the decimal is n – 1.

Thus if the log of a number has a characteristic of –3 then the first two decimal places after the decimal point will be zeros.

Thus, the value will be –3.00ab…

Q.

Find the value of x in 3|3x – 4| = 92x – 2

1.8/7

2.7/8

3.7/4

4.16/7

ANS.1

Explanation :

Take the log of both sides, then we get,

|3x – 4| log 3 = (2x – 2) log 9

= (2x – 2) log 32

= (4x – 4) log 3

Dividing both sides by log 3, we get

|3x – 4| = (4x – 4)

Now, |3x – 4| = 3x – 4 if x > 4/3

so if x > 4/3

3x – 4 = 4x – 4

or 3x = 4x

or 3 = 4

But this is not possible.

Let’s take the case of x < 4/3

Then |3x – 4| = 4 – 3x

Therefore, 4 – 3x = 4x – 4 from

or 7x = 8

or x = 8/7

Q.

Solve for x. log10x – log10√x = 2 logx10

1.101

2.100

3.90

4.80

ANS.2

Explanation :

Now, log10√x = 1/2 * log10x

Therefore, the equation becomes

log10x – 1/2 * log10x = 2 logx10

or 1/2*log10x = 2 logx10

Using base change rule (logba = 1/logab)

Therefore, equation (2) becomes

1/2*log10x = 2/log10x

so (log10x)2 = 4

or log10x = 2

Therefore, x = 100

Q.

If 7x + 1 – 7x – 1 = 48, find x.

1.0

2.1

3.-1

4.2

ANS.1

Explanation :

Take 7x – 1 as the common term. The equation then reduces to

7x – 1(72 – 1) = 48 or 7x – 1 = 1

or x – 1 = 0 or x = 1

Q.

Calculate: log2(2/3) + log4(9/4)

1.0

2.1

3.-1

4.5

ANS.1

Explanation :

= log2(2/3) + (log2(9/4)) /log24

= log2(2/3) + 1/2 log2(9/4)

= log2(2/3) + 1/2 (2 log23/2)

= log22/3 + log23/2 = log21 = 0

Q.

Find the value of the expression 1/log32 + 2/log94 – 3/log278

1.0

2.1

3.-1

4.2

ANS.1

Explanation :

Passing to base 2

we get

log23 + 2log229 – 3log2327

= log2 3 + (4log23)/2 – (9log23)/3 = 3log23 – 3log23

= 0

Q.

Solve the inequality. (a) log2(x + 3) < 2

1.–3 ≤ x < 3

2.–3 ≤ x < 6

3.–3 ≤ x < 9

4.–3 ≤ x < 1

ANS.4

Explanation :

so 22 > x + 3

so 4 > x + 3

1 > x

or x < 1

But log of negative number is not possible.

Therefore, x + 3 ≥ 0

That is, x ≥ –3

Therefore, –3 ≤ x < 1

Q.

Solve the inequality log2(x2 – 5x + 5) > 0

1.x > 3 or x < 1

2.x > 2 or x < 1

3.x > 5 or x < 1

4.x > 4 or x < 1

ANS.4

Explanation :

= x2 – 5x + 5 > 1

= x2 – 5x + 4 > 0

= (x – 4) (x – 1) > 0

Therefore, the value of x will lie outside 1 and 4.

That is, x > 4 or x < 1

Q.

log 32700 = ?

1.log 3.27 + 4

2.log 3.27 + 2

3.2 log 327

4.100 × log 327

ANS.1

Explanation :

Log 32700 = log 3.27 + log 10000 = log 3.27 + 4

Q.

log . 0867 = ?

1.log 8.67 + 2

2.log 8.67 – 2

3.(log 867)/1000

4.–2 log 8.67

ANS.2

Explanation :

Log 0.0867 = log (8.67/100) = log 8.67 – log 100

Log 8.67 – 2

Q.

If log102 = .301 find log10125.

1.2.097

2.2.301

3.2.10

4.2.087

ANS.1

Explanation :

log10 125 = log10(1000/8) = log 1000 – 3log2

= 3 – 3 × 0.301 = 2.097

Q.

log32 8 = ?

1.2/5

2.5/3

3.3/5

4.4/5

ANS.3

Explanation :

log32 8 = log 8/log 32 (By base change rule)

= 3 log2/5log2 = 3/5

Q.

log0.5x = 25; Find the value of x

1.2–25

2.225

3.2–24

4.224

ANS.1

Explanation :

log0.5 x = 25 so x = 0.525 = (½)25 = 2–25

Q.

log3x = 1/2

1.3

2.√3

3.3/2

4.2/3

ANS.2

Explanation :

x = 31/2 = √3

Q.

log153375 × log41024 = ?

1.12

2.14

3.15

4.18

ANS.3

Explanation :

log153375 × log41024

= 3 log15 15 × 5 log4 4 = 3 × 5 = 15

Q.

loga4 + loga16 + loga64 + loga256 = 10. Then a = ?

1.4

2.2

3.8

4.5

ANS.1

Explanation :

The given expression is:

Loga(4 × 16 × 64 × 256) = 10

i.e.loga410 = 10

Thus, a = 4.

Q.

log625√5 = ?

1.4

2.8

3.1/8

4.1/4

ANS.3

Explanation :

1/2 log625 5 = [1/(2 × 4)] log5 5 = 1/8.

Q.

If log x + log (x + 3) = 1 then the value(s) of x will be, the solution of the equation

1.x + x + 3 = 1

2.x + x + 3 = 10

3.x (x + 3) = 10

4.x (x + 3) = 1

ANS.4

Explanation :

log x (x + 3) = 1 so 101 = x2 + 3x.

or x(x + 3) = 10

Q.

If log10a = b, find the value of 103b in terms of a.

1.a3

2.3a

3.a × 1000

4.a × 100

ANS.1

Explanation :

log10a = b so 10b = a so By definition of logs.

Thus 103b = (10b)3 = a3

Q.

3 log 5 + 2 log 4 – log 2 = ?

1.4

2.3

3.200

4.1000

ANS.2

Explanation :

3 log 5 + 2 log 4 – log 2

= log 125 + log 16 – log 2

= log (125 × 16)/2 = log 1000 = 3

Q.

log (3x – 2) = 1; the value of x

1.3

2.2

3.4

4.6

ANS.3

Explanation :

101 = 3x – 2 so x = 4.

Q.

log (2x – 3) = 2; so x=?

1.103

2.51.5

3.25.75

4.26

Explanation :

102 = 2x – 3 so x = 51.5

Q.

log (12 – x) = –1

1.11.6

2.12.1

3.11

4.11.9

ANS.4

Explanation :

1/10 = 12 – x so x = 11.9