LOGARITHMS Ch 1-1
1 Logarithms
KEY FACTS
1. Definition: If a and n are positive real numbers such that a ≠ 1 and x is real, then ax = n ⇒ x = logan.
Here x is said to be the logarithm of the number n to the base a.
1
Ex. 43 = 64⇒ log4 64 = 3, 10–1 = = 0.1 ⇒ log10 0.1 = – 1, 5 x = 4 ⇒ x = log54,
10
a0 = 1 ⇒ loga 1 = 0, a1 = a ⇒ loga a = 1.
2. Some Important Facts about Logarithms
● logan is real if n > 0
● logan is imaginary if n < 0
● logan is not defined if n = 0
● The logarithm of 1 to any base a, a > 0 and a ≠ 1 is zero. log a 1 = 0
● The logarithm of any number a, a > 0 and a ≠ 1, to the same base is 1. log a a = 1
log x
● If a and x are positive real numbers, where a ≠ 1, then a a = x
Proof. Let logax = p. Then, x = a p (By def.) ⇒ x = alogax (Substituting the value of p)
Ex. 3log3 7 = 7, 2log2 9 = 9, 5log5 x = x
● For a > 0, a ≠ 1, logax1 = logax2 ⇒ x1 = x2 (x1, x2 > 0)
● If a > 1 and x > y, then logax > logay.
● If 0 < a < 1 and x > y, then logax < logay
3. Laws of Logarithms
For x > 0, y > 0 and a > 0 and a ≠ 1, any real number n
● logaxy = logax + logay Ex. log2(15) = log2(5 × 3) = log25 + log23
3
● loga(x/y) = logax – logay Ex. log 2 = log 2 3 − log 2 7
7
n
● loga(x) = n logax Ex. log (2)5 = 5 log 2,
a3 3 3
log 3 = log a – log b = 3 log a – 3 log b
b
1 1
● logax = Ex. log5 2 =
log x a log 2 5
1 1 1
● log a n x = log a x Ex. log87 = log23(7) = log 2 7, log 5 3 = log (5) 1 (3) = log51/2 (3) = log5 3 = 2log5 3
n 3 2 1/2
m m 4 4
● log a n x = log a x Ex. log 25 5 = log 2 5
n 5
Ch 1-1
,Ch 1-2 IIT FOUNDATION MATHEMATICS CLASS – IX
Base changing formula
● logax = logbx.logab
Ex. log12 32 = log16 32. log1216. (The base has been changed from 12 to 16)
↑ ↑
Old base New base
● xlogay = ylogax Ex. 3log 7 = 7log 3 (It being understood that base is same)
y→
[Proof. x log a = x
log x y . log a x
(Base changing formula)
log x
= ( x log x y )
a
(Using n loga x = (loga x)n)
log a x
= y (Using x logx y = y.)
log b
● logab = (It being understood that base is same)
log a
● If logab = x for all a > 0, a ≠ 1, b > 0 and x ∈ R, then log1/a b = – x, loga 1/b = – x and log1/a 1/b = x
4. Some Important Properties of Logarithms
● a, b, c are in G.P. ⇔ logax, logbx, logcx are in H.P.
● a, b, c are in G.P. ⇔ logxa, logxb, logxc are in A.P.
5. Natural or Naperian logarithm is denoted by logeN, where the base is e.
1
Ex. loge7, loge , logeb, etc.
64
● Common or Brigg’s logarithm is denoted by log10N, where the base is 10.
1
Ex. log105, log10 , etc.
81
● logax is a decreasing function if 0 < a < 1
● logax is an increasing function if a > 1.
6. Characteristic and Mantissa
● Characteristic: The integral part of the logarithm is called characteristic.
(i) If the number is greater than unity and there are n digits in integral part, then its characteristic = (n – 1)
(ii) When the number is less than 1, the characteristic is one more than the number of zeroes between the decimal
point and the first significant digit of the number and is negative. It is written as (n + 1) or Bar (n + 1).
Ex. Number Characteristic Number Characteristic
4.1456 0 0.823 1
24.8920 1 0.0234 2
238.1008 2 0.000423 4
7. Arithmetic Progression. A sequence a1, a2, a3, ........, an is said to be in arithmetic progression, when
a2 – a1 = a3 – a2 = .......... = an – an – 1, i.e., when the terms in the sequence increase or decrease by a constant
quantity called the common difference.
Ex. 1, 3, 5, 7, 9, ....
6, 11, 17, 23, .... – 5, –2, 1, 4, 7, ....
● Sum of first ‘n’ terms of an Arithmetic Progression
n n
Sn = [2a + ( n – 1)d ] = [a + l ],
2 2
where a = first term, n = number of terms, d = common difference, l = last term.
● Sum of first “n” natural numbers.
n( n + 1)
Sn = 1 + 2 + 3 + ............ + n = .
2
, LOGARITHMS Ch 1-3
n( n + 1)
Also, written as Sn =
2
● Also, if a, b, c are in A.P. then 2b = a + c
8. Geometric Progression : A sequence a1, a2, a3, ............, an is said to be in Geometric Progression when,
a2 a a a
= 3 = 4 = ............ = n = r (say)
a1 a2 a3 an – 1
where a1, a2, a3, .......... are all non zero numbers and r is called the common ratio.
Ex. 3, 6, 12, 24, .................. r = 2;
1 1 1 1
64, 16, 4, 1, , . , ............ r =
4 16 64 4
a ( r n – 1) a (1 – r n ) lr – a
● Sum of first n terms of a G.P. Sn = if r > 1 = if r < 1 =
( r – 1) (1 – r ) r –1
where, a = first term, r = common ratio, l = last term
a
● Sum of an infinite G.P. S∞ = , where a = first term, r = common ratio.
1–r
● For three terms a, b, c to be in G.P., b2 = ac
9. Harmonic Progression : A series of quantities a1, a2, a3, ............, an are said to be in H.P. when their reciprocals
1 1 1 1
, , , .........., are in A.P.
a1 a2 a3 an
2ac
● When three quantities a, b, c are in H.P., then, b = .
a+c
SOLVED EXAMPLES
Ex. 1. If loga5 + loga25 + loga125 + loga625 = 10, then find the value of a.
Sol. loga5 + loga25 + loga125 + loga625 = 10
⇒ loga (5 × 25 × 125 × 625) = 10
⇒ loga (51 × 52 × 53 × 54) = 10
⇒ loga 510 = 10 ⇒ a10 = 510 ⇒ a = 5.
[Using loga x = n ⇒ x = an]
Ex. 2. Solve for x : log10 [log2 (log39)] = x.
Sol. log10 [log2 (log39)] = x
⇒ log2 (log39) = 10x
⇒ log2 (log3 32) = 10x
⇒ log2 (2 log33) = 10x
⇒ log2 2 = 10x ⇒ 10x = 1 = 100 ⇒ x = 0.
3 5 2 n −1
Ex. 3. Find the value of logxx + log x x + log x x + ........ + log x x .
3 5 2n − 1
Sol. logxx + log x x + log x x + ........ + log x x = log x x + 3 log x x + 5 log x x + ......... + (2n – 1) log x x
n 2 n
= 1 + 3 + 5 + ............ + (2n – 1) = [1 + (2n – 1)] = n Using log x x = 1 and for A.P. Sn = 2 (a + l )
2
1 Logarithms
KEY FACTS
1. Definition: If a and n are positive real numbers such that a ≠ 1 and x is real, then ax = n ⇒ x = logan.
Here x is said to be the logarithm of the number n to the base a.
1
Ex. 43 = 64⇒ log4 64 = 3, 10–1 = = 0.1 ⇒ log10 0.1 = – 1, 5 x = 4 ⇒ x = log54,
10
a0 = 1 ⇒ loga 1 = 0, a1 = a ⇒ loga a = 1.
2. Some Important Facts about Logarithms
● logan is real if n > 0
● logan is imaginary if n < 0
● logan is not defined if n = 0
● The logarithm of 1 to any base a, a > 0 and a ≠ 1 is zero. log a 1 = 0
● The logarithm of any number a, a > 0 and a ≠ 1, to the same base is 1. log a a = 1
log x
● If a and x are positive real numbers, where a ≠ 1, then a a = x
Proof. Let logax = p. Then, x = a p (By def.) ⇒ x = alogax (Substituting the value of p)
Ex. 3log3 7 = 7, 2log2 9 = 9, 5log5 x = x
● For a > 0, a ≠ 1, logax1 = logax2 ⇒ x1 = x2 (x1, x2 > 0)
● If a > 1 and x > y, then logax > logay.
● If 0 < a < 1 and x > y, then logax < logay
3. Laws of Logarithms
For x > 0, y > 0 and a > 0 and a ≠ 1, any real number n
● logaxy = logax + logay Ex. log2(15) = log2(5 × 3) = log25 + log23
3
● loga(x/y) = logax – logay Ex. log 2 = log 2 3 − log 2 7
7
n
● loga(x) = n logax Ex. log (2)5 = 5 log 2,
a3 3 3
log 3 = log a – log b = 3 log a – 3 log b
b
1 1
● logax = Ex. log5 2 =
log x a log 2 5
1 1 1
● log a n x = log a x Ex. log87 = log23(7) = log 2 7, log 5 3 = log (5) 1 (3) = log51/2 (3) = log5 3 = 2log5 3
n 3 2 1/2
m m 4 4
● log a n x = log a x Ex. log 25 5 = log 2 5
n 5
Ch 1-1
,Ch 1-2 IIT FOUNDATION MATHEMATICS CLASS – IX
Base changing formula
● logax = logbx.logab
Ex. log12 32 = log16 32. log1216. (The base has been changed from 12 to 16)
↑ ↑
Old base New base
● xlogay = ylogax Ex. 3log 7 = 7log 3 (It being understood that base is same)
y→
[Proof. x log a = x
log x y . log a x
(Base changing formula)
log x
= ( x log x y )
a
(Using n loga x = (loga x)n)
log a x
= y (Using x logx y = y.)
log b
● logab = (It being understood that base is same)
log a
● If logab = x for all a > 0, a ≠ 1, b > 0 and x ∈ R, then log1/a b = – x, loga 1/b = – x and log1/a 1/b = x
4. Some Important Properties of Logarithms
● a, b, c are in G.P. ⇔ logax, logbx, logcx are in H.P.
● a, b, c are in G.P. ⇔ logxa, logxb, logxc are in A.P.
5. Natural or Naperian logarithm is denoted by logeN, where the base is e.
1
Ex. loge7, loge , logeb, etc.
64
● Common or Brigg’s logarithm is denoted by log10N, where the base is 10.
1
Ex. log105, log10 , etc.
81
● logax is a decreasing function if 0 < a < 1
● logax is an increasing function if a > 1.
6. Characteristic and Mantissa
● Characteristic: The integral part of the logarithm is called characteristic.
(i) If the number is greater than unity and there are n digits in integral part, then its characteristic = (n – 1)
(ii) When the number is less than 1, the characteristic is one more than the number of zeroes between the decimal
point and the first significant digit of the number and is negative. It is written as (n + 1) or Bar (n + 1).
Ex. Number Characteristic Number Characteristic
4.1456 0 0.823 1
24.8920 1 0.0234 2
238.1008 2 0.000423 4
7. Arithmetic Progression. A sequence a1, a2, a3, ........, an is said to be in arithmetic progression, when
a2 – a1 = a3 – a2 = .......... = an – an – 1, i.e., when the terms in the sequence increase or decrease by a constant
quantity called the common difference.
Ex. 1, 3, 5, 7, 9, ....
6, 11, 17, 23, .... – 5, –2, 1, 4, 7, ....
● Sum of first ‘n’ terms of an Arithmetic Progression
n n
Sn = [2a + ( n – 1)d ] = [a + l ],
2 2
where a = first term, n = number of terms, d = common difference, l = last term.
● Sum of first “n” natural numbers.
n( n + 1)
Sn = 1 + 2 + 3 + ............ + n = .
2
, LOGARITHMS Ch 1-3
n( n + 1)
Also, written as Sn =
2
● Also, if a, b, c are in A.P. then 2b = a + c
8. Geometric Progression : A sequence a1, a2, a3, ............, an is said to be in Geometric Progression when,
a2 a a a
= 3 = 4 = ............ = n = r (say)
a1 a2 a3 an – 1
where a1, a2, a3, .......... are all non zero numbers and r is called the common ratio.
Ex. 3, 6, 12, 24, .................. r = 2;
1 1 1 1
64, 16, 4, 1, , . , ............ r =
4 16 64 4
a ( r n – 1) a (1 – r n ) lr – a
● Sum of first n terms of a G.P. Sn = if r > 1 = if r < 1 =
( r – 1) (1 – r ) r –1
where, a = first term, r = common ratio, l = last term
a
● Sum of an infinite G.P. S∞ = , where a = first term, r = common ratio.
1–r
● For three terms a, b, c to be in G.P., b2 = ac
9. Harmonic Progression : A series of quantities a1, a2, a3, ............, an are said to be in H.P. when their reciprocals
1 1 1 1
, , , .........., are in A.P.
a1 a2 a3 an
2ac
● When three quantities a, b, c are in H.P., then, b = .
a+c
SOLVED EXAMPLES
Ex. 1. If loga5 + loga25 + loga125 + loga625 = 10, then find the value of a.
Sol. loga5 + loga25 + loga125 + loga625 = 10
⇒ loga (5 × 25 × 125 × 625) = 10
⇒ loga (51 × 52 × 53 × 54) = 10
⇒ loga 510 = 10 ⇒ a10 = 510 ⇒ a = 5.
[Using loga x = n ⇒ x = an]
Ex. 2. Solve for x : log10 [log2 (log39)] = x.
Sol. log10 [log2 (log39)] = x
⇒ log2 (log39) = 10x
⇒ log2 (log3 32) = 10x
⇒ log2 (2 log33) = 10x
⇒ log2 2 = 10x ⇒ 10x = 1 = 100 ⇒ x = 0.
3 5 2 n −1
Ex. 3. Find the value of logxx + log x x + log x x + ........ + log x x .
3 5 2n − 1
Sol. logxx + log x x + log x x + ........ + log x x = log x x + 3 log x x + 5 log x x + ......... + (2n – 1) log x x
n 2 n
= 1 + 3 + 5 + ............ + (2n – 1) = [1 + (2n – 1)] = n Using log x x = 1 and for A.P. Sn = 2 (a + l )
2