Let a be a positive real number, a ≠ 1 and ax = m. Then x is called the logarithm of m to the base a and is written as logam, and conversely, if logam = x, then ax = 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, logam = x, and we can say that m > 0 and a > 0.
Thus ax = m so x = logam and logam = x so ax = m
In short, ax = m so x = logam.
x = logam is called the logarithmic form and ax = m is called the exponential form of the equation connecting a, x and m.
Two Properties of Logarithms :
1. loga1 = 0 for all a > 0, a ≠ 1 That is, log 1 to any base is zero
2. logaa = 1 for all a > 0, a ≠ 1 That is, log of a number to the same base is 1
Laws of Logarithms :
First Law: loga(mn) = logam + logan That is, log of product = sum of logs
Second Law: loga(m/n) = logam – logan 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: logamn = n logam
Generalisation
log (mnp) = log m + log n + log p
\( log (a_1a_2a_3 ... a_k) = log a_1 + log a_2 + ... + log a_k \)
Note: Common logarithms: We shall assume that the base a = 10 whenever it is not indicated. Therefore, we shall denote log10m 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) loga(b) = logc(b)/logc(a)
It is one of the most important rules for solving logarithms. (ii) logb(a) = logc(a)/logb(c)
A corollary of this rule is (iii) loga(b) = 1/logb(a)
(iv) log c to the base ab is equal to \( \frac{log a^c}{b} \).
Results on Logarithmic Inequalities :
(a) If a > 1 and loga x1 > loga x2 then x1 > x2
(b) If a < 1 and loga x1 > loga x2 then x1 < x2
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 an + 1 – an.
Example: log10 x = 2.bcde..., then the number of integral values that x can take is given by: 102 + 1– 102 = 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 log10 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...
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
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
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
= 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
Passing to base 2 we get log23 + 2log229 – 3log2327 = log2 3 + (4log23)/2 - (9log23)/3 = 3log23 – 3log23 = 0
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
= 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
Log 32700 = log 3.27 + log 10000 = log 3.27 + 4
Log 0.0867 = log (8.67/100) = log 8.67 – log 100 Log 8.67 – 2
log10 125 = log10(1000/8) = log 1000 – 3log2 = 3 – 3 × 0.301 = 2.097
log32 8 = log 8/log 32 (By base change rule) = 3 log2/5log2 = 3/5
log0.5 x = 25 so x = 0.525 = (½)25 = 2–25
x = 31/2 = √3
log153375 × log41024 = 3 log15 15 × 5 log4 4 = 3 × 5 = 15
The given expression is: Loga(4 × 16 × 64 × 256) = 10 i.e.loga410 = 10 Thus, a = 4.
1/2 log625 5 = [1/(2 × 4)] log5 5 = 1/8.
log x (x + 3) = 1 so 101 = x2 + 3x. or x(x + 3) = 10
log10a = b so 10b = a so By definition of logs. Thus 103b = (10b)3 = a3
3 log 5 + 2 log 4 – log 2 = log 125 + log 16 – log 2 = log (125 × 16)/2 = log 1000 = 3
101 = 3x – 2 so x = 4.
102 = 2x – 3 so x = 51.5
1/10 = 12 – x so x = 11.9
x2 – 6x + 6 = 100 so x2 – 6x + 6 = 1 so x2 – 6x + 5 = 0 Solving gives us x = 5 and 1
x log 2 = 3 log 2 = 3/x. Therefore, x = 3/log 2
3x = 7 so log3 7 = x Hence x = 1/log7 3
x = log5 10 = 1/log10 5 = 1/log 5.
x = log0.01 2 = –log 2/2
log x = log (7.2/2.4) = log 3 fi x = 3
log x = log 18 fi x = 18
log x = log 25 + log 8 = log (25 × 8 ) = log 200.
log (x –13) + log 8 = log[3x + 1] so log (8x – 104) = log (3x + 1) so 8x – 104 = 3x + 1 5x = 105 so x = 21
log (2x – 2)/(11.66 – x) = log 30 so (2x – 2)/(11.66 – x) = 30 2x – 2 = 350 – 30x Hence, 32x = 352 so x = 11.