Logarithm

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Logarithm
Meaning of Logarithm:
Logarithm is a rule which is used to solve complex algebraic expression.
Exponential Form:
Every positive real number can be expressed as N = ax, a > 0, a ≠ 1, N>0
a → base, x → exponent
Q.1 Express 16 using exponential form with base 2, base 4 and base 16.
Sol. (i)N = 16, a = 2, Suppose x = power
1
2
x1
(ii)
1
= 16 ⇒ 2 = 16
4
a2 = 4, power = x2
4x2 = 16 ⇒ 42 = 16
(iii) a3 = 16, power = x3
16x3 = 16 ⇒ 161 = 16
Q.2
Sol.
Express 81 using exponential form with base 3, base 9, and base 81.
(i)
N = 81, a1 = 3, x1 = Power
3x1 = 81 ⇒ 34 = 81
(ii)
a2 = 9, x2 = Power
9x2 = 81 ⇒ 92 = 81
(iii) a3 = 81, x3 = Power
81x3 = 81 ⇒ 811 = 81
Express 16 using exponential form with base ‘3’.
It means, 3x = 16
We know that, 32 = 9 and 33 = 27
⇒ 2 < x < 3 ⇒ x ∈(2, 3)
Note:
(i)
What if a = 1.
Express 16 in exponential form with
base a = 1
1x = 16 ⇒ there does not exist any ‘x’.
For which 1x = 16
So we can not take base a = 1.
The types of ‘x’ can be find out
using logarithm.
Ex.
1
Suppose a = –2 and x= − then
2
(−2)
−
1
2
does not give real number.
1.
Logarithm
Q.3
Sol.
(ii) What if a is negative.
Then for even root it will not be defined.
Hence for negative base it is not uniquely
defined that with powers it will give real
number or not.
Logarithmic form:
Logarithm of a number to some base is the
exponent by which the base must be raised
in order to get that number.
logaN = x ⇔ ax = N, N > 0, a > 0, a ≠ 1
a → base, x → exponent, N → number
Q.1
Sol.
Q.
Sol.
Find values of following:
(i) log101000
Suppose log101000 = x ⇔ 10x = 1000
⇒ 10x = 103 (by comparing)
⇒ x = 3
(ii) log232
Suppose log232 = y ⇔ 2y = 32
⇒ 2y = 25 (by comparing)
⇒ y = 5
Q.
Sol.
(iii) log5625
Q.
(iv) log 2 64
Sol.
Suppose log 2 64 = y ⇔
Suppose log5625 = x ⇔ 5x = 625
⇒ 5x = 54 (by comparing)
⇒ x = 4
( 2)
y
= 64
y
Logarithm
⇒ 22 = 26
y
⇒
= 6 (by comparing)
2
⇒ y = 12
2.
Q.
Sol.
(v) log 2 3 1728
( )
Suppose log 1728 = x ⇔ 2 3
2 3
( )
⇒ 2 3
x
= 1728 = 123
( )
2


=  2 3 


x
( ) ( )
⇒ 2 3
Q.
Sol.
Q.
Sol.
x
= 2 3
6
3
(by comparing)
⇒x=6
(vi) log1632
Suppose log1632 = y ⇔ 16y = 32
⇒ (24)y = 25
⇒ 24y = 25
⇒ 4y = 5 (by comparing)
5
⇒ y =
4
(vii) log 3 7 2401
Suppose log 3 7 2401 = x ⇔
( 7)
3
x
= 2401
x
⇒ 7 3 = 74
x
⇒
= 4 (by comparing)
3
⇒ x = 12
Sol.
Q.2
(viii) log 3 (log 3 (27)3 )
First we will find the value of log3(27)3 = y
⇒ log3(33)3 = y
⇒ log339 = y
⇒y=9
Then it reduces to log39 and we know that log39 = 2
Hence, log3(log3(27)3) = 2
Find ‘x’ for which log(x – 1) (2x + 1) is defined.
 –1 
(A) x ∈ (1, ∞ )(B) x ∈  , ∞ – {2}
2 
(C) x∈ (1, ∞ ) – {2}
 –1 
(D) x ∈  , 1
2 
3.
Logarithm
Q.
Sol. (C)
We will check when log(x – 1)(2x + 1) is defined.
x – 1 > 0 and (x – 1) ≠ 1 and (2x + 1) > 0
–1
⇒ x > 1 and x ≠ 2 and x >
2
 –1 
 2 , ∞
Intersection gives x ∈ (1, ∞ ) and x ≠ 2 ⇒ x ∈ (1, ∞ ) – {2}
It can also be written as
x ∈ (1, 2) ∪ (2, ∞)
⇒ x ∈ (1, ∞ ) and x ≠ 2 and x ∈
Q.3 Find complete set of real values of ‘x’ for which log
Sol. 2x – 3 > 0 and (2x – 3) ≠ 1 and (x – 5x – 6) > 0
(x2–5x–6) is defined.
(2x–3)
2
3
and x ≠ 2 and (x – 6) (x + 1) > 0
2
3
⇒ x>
and x ≠ 2 and x ∈ ( −∞, −1) ∪ (6, ∞ )
2
⇒ x>
3 
⇒ x ∈  , ∞  and x ≠ 2 and x ∈ ( −∞, −1) ∪ (6, ∞ )
2 
Intersection gives, x ∈ (6, ∞ )
Q.4
Let
a = log1/216
b = log3(tan30°)
c = log 2− 3 2 + 3
(
)
d = log2(log24)
then a + b + c + d is
7
9
5
(A) − (B) − (C) − 2
2
2
(D) –4
Sol. (B)
a = log1/216
a
1
⇒   = 16
2
⇒ 2–a = 24
⇒a=–4
b = log3(tan30°)
Logarithm
⇒ 3b = tan30° =
⇒ b= −
1
−
1
=32
3
1
2
4.
(2 + 3 )
(2 + 3 )(2 − 3 )
3) =
(2 − 3 )
3 ) = ( 4 − 3) (2 − 3 )
c = log 2−
(
⇒ 2−
(
⇒ 2−
3
c
c
c = –1
d = log2(log24)
we know that log24 = 2
so, d= log22 ⇒ d = 1
then a + b + c + d = −4 −
−1
1
9
−1+1 = −
2
2
Important Deduction:
(i) logNN = 1
(ii) logN1 = 0
(iii) log1/N N = –1
(i) Proof : Suppose logNN = x ⇔ Nx = N 1 ⇒ x = 1
(ii) Proof : Suppose logN1 = y ⇔ Ny = 1 = N0 ⇒ y = 0
z
= N ⇒ N−z = N1 ⇒ z = –1
Q.5
Find values of following
(i) logsin30°cos60°
Sol.
1
First simplify this, we get log 1  
2 2
Know the facts
1
so log 1   = 1
2 2
We know that logNN =1
Q.
(ii) log4/3 1.3
Sol.
Let
N = 1.3 = 1.3333….
N = 1.3333…
10 N = 13.33333.......
...(i)
...(ii)
Know the facts
We know that logNN =1
5.
Logarithm
 1
(iii) Proof : Suppose log 1 N = z ⇔  
 N
N
Subtracting (i) from (ii), we get
– 9N = –12
–12 4
N=
=
–9 3
 4
We get log 4   = 1
3 3
Q.
Sol.
(iii)
log 5 5 5 5 . . .
5
5 . . .
Sol. N = 5
N
N= 5N
N2 = 5N
⇒ N2 –5N = 0
⇒ N(N–5) = 0
⇒ N = 0 or N = 5 but N = 0 is not possible
We get log55 = 1
Q.
Sol.
Q.
Sol.
(iv) log2(sin2x + cos2x)
We know that sin2x + cos2 x = 1
So we get , log21 = 0
(v)
 log 1 = 0
N
log(tan1°).log(tan2°).log(tan3°)…log(tan89°)
log(tan1°). log(tan2°)… log(tan45°)…log(tan89°)
= log(tan1°). log(tan2°)… log1…log(tan89°)
=0
Q. (vi) log(tan1°.tan2°.tan3°... tan89°)
Sol. We get
Logarithm
log(tan1°·tan2°·tan3°...tan43°·tan44°·tan45°·tan46°...tan89°)
= log(tan1°·tan2°... tan44°·tan45°·tan(99°–44°)...tan(90–1)°)
= log(tan1°·tan2°... tan44°·1·cot44°·cot43°...cot1°)
= log(tan1°·cot1°·tan2°cot2°...tan44°cot44°)
= log1 = 0
Using tan(90°–θ) = cotθ
and tanθ.cotθ = 1
6.
Q. (vii) log (log (log (27) ))
Sol. Let’s assume
2
3
3
3
log3(27)3 = y
⇒ log3(33)3 = y
⇒ log339 = y
⇒y=9
Then it reduces to log 2 (log 3 9) = log 2 2 = 1

2
Q.
Sol.
(log
(viii)
100
(
)
(
10 ) log 2 (log 4 2 ) log 4 log 22 ( 256 )
log 4 8 + log 8 4
2
)
Let a = log10010
⇒ 100a = 10
⇒ (10)2a = 101
⇒ 2a = 1
1
⇒a=
2
b = log2(log42)
⇒ b = log 2
1
2
⇒ 2b = 2–1
⇒b=–1
2
2
2
2
Let x = log 22 ( 256 ) ⇒ x = log 2 ( 28 )  ⇒ x = (log 2 216 )


⇒ x = 162 = 256
(
c = log 4 log
2
2
2
(256)
)
So we get c = log4256
⇒ 4c = 44
⇒c=4
d = log48
⇒ 4d = 8
⇒ (22)d = 23
⇒ 22d = 23
⇒ 2d = 3
⇒d= 3
2
e = log84
⇒ 8e = 4
⇒ (23)e = 22
⇒ 3e = 2
2
⇒e=
3
1
  ( −1) ( 4 )
( −2)·6
abc
12
2
= 
=
=−
Then given expansion reduces to
3 2
d+e
13
(9 + 4)
+
2 3
Q.6 If log (log (log x)) = 0 = log (log (log y)) then find x + y.
Sol. log (log (log x) = 0 ⇔ log (log x) = 2 = 1 (Change it into exponential form)
2
2
2
3
3
2
2
3
3
1
⇔ log3x = 2 = 2
⇔ x = 32 = 9
x=9
2
0
7.
Logarithm
2
Similarly, log2(log3(log2y) = 0 ⇔ log3(log2y) = 20 = 1
(Change it into exponential form)
⇔ log2y = 31 = 3
⇔ y = 23 = 8
Then x + y = 9 + 8 = 17
Fundamental Identify:
Proof:
alogaN = N
Let logaN = x then it changes into
ax = N ⇔ logaN = x
So we get alogaN = N
Q.
Sol.
Find the value of following: 3log3 10
Using Identify, alogaN = N
We get 3log3 10 = 10
Properties of logarithm:
If m, n are positive real numbers, a > 0, a ≠ 1 then
(i)
logamn = logam + logan
Proof:
Suppose
logam = x and logan =y
ax = m and ay = n
then
mn = ax.ay = ax + y
again change it into logarithmic form, x + y = logamn
we get logamn = logam + logan
Logarithm
Q.1
Sol.
Solve : log102 + log105
log102 + log105
= log10(2×5)
= log1010
=1
(using P(1))
8.
Note:
General version:
Suppose a > 0, a ≠ 1, N1, N2, N3, …, Nr > 0
loga(N1.N2.N3 ... Nr) = logaN1 + logaN2 + logaN3 + …. + logaNr
(ii)
m
log a   = logam – logan
n
Proof:
Given a > 0, a ≠ 1, m > 0, n > 0
Then suppose logam = x & logan = y
⇒
ax = m & ay = n
m ax
⇒
=
= ax – y
n ay
Change it into logarithmic form,
m
x – y = log a  
n
m
⇒ logam – logan = log a  
n
Q.2 Solve : log 10 – log 5
Sol. log 10 – log 5 = log 10
5
2
2
2
2
2
(using P(2))
= log22
= 1
logamn = n logam
Proof:
Let’s say logam = α ⇔ aα = m
then mn = (aα)n = anα
Change it into logarithmic form, then
nα = logamn
nlogam = logamn
9.
Logarithm
(iii)
Q.3 Q3. Solve log 32
Sol. log 32 = log 2
2
2
Q.4
5
2
= 5log22
=5
Find values of following :
(i) log3855 + log3857 + log38511
Sol. Using property log m + log n + log p = log mnp
a
a
We get log3855.7.11
= log385385
= 1
a
a
Q. (ii) log (log (log 625)))
Sol. log (log (log 5 ))
4
4
=
=
=
=
=
2
2
5
5
4
log4(log2(4log55))
log4(log2(22 × 1))
log4(2log22)
log42
1/2
Q. (iii) log (11)log
Sol. log (11)log 1331
11
11 1331
11
11
3
= log 11 11log11 11
[Using logamn = nlogam]
= log 11 113log11 11 = log11113
= 3log1111
=3
Q. (iv) log 10 – log 5 + log 8
Sol. log 10 – log 5 + log 8
2
Logarithm
2
2
2
2
2
10
= log 2
+ log 2 8 5
= log22 + log223
= 1 + 3log22
=1+3
=4
[Using P(1)]
10.
Q.
Sol.
Q.5
Sol.
(v)
log2[log4(log10164 + log10258)]
log2[log4(log10164 + log10258)]
= log2[log4(log10164.258)]
= log2[log4(log1024×4.52×8)]
= log2[log4(log10(10)16)]
= log2[log4(16log1010)]
= log2[log442]
= log22(log44)
= log22 = 1

1
 1 +  is equal to
n
n= 1

(A) 8
(B) 9
1023
∑ log
2
(C)
1023
∑ log
n= 1
2
(C) 10 (D) 12
n + 1


 n 
2
3
4
 1024 
= log 2   + log 2   + log 3   + ... + log 2 

1
2
3
 1023 
 2 3 4 1024 
= log 2  . . ...

 1 2 3 1023 
= log21024
= log2210
= 10log22 = 10
Sol.
 16 
 25 
 81 
Find the value of log 10 2 + 16log 10   + 12log 10 
 + 7 log 10 
.
 15 
 24 
 80 
log102 + 16log1016 – log1015 + 12log1025 – 12log1024 + 7log1081 – 7log1080
= log102 + 16log1024 –16log10(3×5)– log10(3×5) + 12log1052 – 12log10(23.3)
+ 7log1034 – 7log10(24.5
= log102 + 64log102 – log103 – log35 + 24log105 – 36log102
– 12log103 + 28log103 – 28log102 – 7log105
= (1 + 64 – 36 – 28) log102 + (–16 – 12 + 28) log103 + (–16 – 24 – 7) log105
= log102 + log105
= log1010 = 1
11.
Logarithm
Q.6
Base Changing theorem:
log c a
, a > 0, b > 0, c > 0, a ≠ 1, b ≠ 1, c ≠ 1
log c b
logba =
Proof:
Suppose logba = x ⇔ a = bx
logca = logcbx
logca = xlogcb
log c a
= x = logba
log c b
⇒ logb a =
(By taking logarithm)
log c a
log c b
Note:
log a b =
1
logb a
Proof:
log c b
1
=
log
log c a
c a
log c b
(By using base Changing Theorem)
log c b log c b
=
log c a log c a
(Both are equal to each other)
Hence proved.
Q.7
Prove the following: logba . logcb . logdc = logda
Sol. log a . log b . log c =
b
c
Logarithm
= logda
d
log a log b log c log a
.
.
=
log b log c log d log d
(Using Base-Changing Theorem)
12.
Q.8
Sol.
If log23. log34 . log45...logn(n + 1) = 10. Find ‘n’.
log23. log34 . log45...logn(n + 1) = 10
log 3 log 4 log 5
log(n + 1)
.
.
. ... .
= 10
log 2 log 3 log 4
log n
log(n + 1)
= 10 log 2
⇒
(Using Base-Changing Theorem)
⇒ log2(n + 1) = 10
⇒ 210 = n + 1
⇒ n = 1024 –1 ⇒ n = 1023
Property of logarithm:
alogbc = clogba
Proof:
a
logbc
=a
log a c
log ab
1
= ( alogac )logab
= (c )
1
log ab
(Using Base-Changing theorem)
= clogba
Hence alogbc = clogba
Property of logarithm:
( )
log ak mn =
n
log a m
k
wherever defined
Proof:
L.H.S.=
log cmn
log c a
k
=
nlog cm
klog c a
=
n
log a m = R.H.S.
k
(Using Base-changing theorem)
Hence Proved.
The value of 7log3 5 + 3log5 7 − 5log3 7 − 7log5 3 is equal to
(A) 3
(B) 5
(C) 7 Sol.
(D)
7log3 5 + 7log5 3 − 7log3 5 − 7log5 3 = 0
(D) 0
(Using property a log cb = blogc a )
13.
Logarithm
Q.9
Q.10 Let x
1
= log1632, x2 = log625125, x3 = log816, x4 = log 2 8 . Find
4
∑x .
i= 1
i
5
5
log 2 2 =
4
4
3
3
x2 = log 54 53 = log 5 5 =
4
4
4
4
x3 = log 23 24 = log 2 2 =
3
3
Sol. x
1
= log 24 25 =
x4 = log 21/2 23 = 6log 2 2 = 6
4
∑x
then
i
i= 1
= x1 + x2 + x3 + x4 =
log 3 7
Q.11 Let
1
A= 
9
Sol. A = (3 )
−2
log 3 7
−1
, B = 2log1/2 7 , C = 8log3 2 , D = 4−log2 6 . Find
( )−2
log 3 7
=3
5 3 4
28
+ + +6 =
4 4 3
3
1 1 1 1
.
+ − −
A B C D
1
49
=
1
7
−3
1
C = 23( −log2 3) = 2log2 3 =
27
−2
1
D = 2−2log2 6 = 2log2 6 =
36
1 1 1 1
= 49 + 7 – 27 – 36 = –7
+ − −
A B C D
log
B=2
Q.12 If p
log 3 7
Sol. p
(2−1 )
7
−1
= 2log2 7 =
2
= 81 , then find value of p(log3 7) .
log 3 7.log 3 7
= (plog3 7 )
log 3 7
log 3 7
= ( 81)
4
= 34log3 7 = 3log3 7 = 74 = 2401
Q.13 If a, b, c are real positive numbers such that
(log 7 11)2
(log 3 7 )2
then find the value of a
+b
(A) 343
(B) 121
a
log 3 7
log 7 11
= 27, b
log 11 25
= 49, c
=
11 ,
(log 11 25)2
+c
.
(C) 469 (D) 569
Sol. (C)
alog3 7.log3 7 + blog 7 11.log 7 11 + clog11 25.log11 25
= ( alog3 7 )
Logarithm
log 3 7
log 3 7
= ( 27 )
+ (blog 7 11 )
log 7 11
log 7 11
+ ( 49)
+
+ ( clog11 25 )
log 11 25
( 11 )
log 11 25
14.
= 33log3 7 + 72log 7 11 + 11½log11 25
= 3log3 7 + 7log 7 11 + 11log11 (25)
= 73 + 112 + 5
= 343 + 121 + 5 = 469
3
2
1/2
Q.14 If log x = b for permissible values of a and x then identify the statement(s)
a
which can be correct?
(A) If a and b are two irrational numbers then x can be rational.
(B) If a rational and b irrational then x can be rational.
(C) If a irrational and b rational then x can be rational.
(D) If a rational and b rational then x can be rational.
Sol.
(ABCD)
Change this into exponential form
(A) x = ab
Suppose a = 2, b = log 2 3
First we will prove that log23 is not rational
Suppose log23 is rational, then log23 = p/q ⇒ 3 = 2p/q
⇒ 3q = 2p
It is not possible for any value of p, q except p = 0 and q = 0
So, log23 is irrational.
1
Hence log 2 3 is irrational
2
So, we can say log 2 3 will be irrational
Then
( 2)
log
2
3
= 3 is rational.
(B) a = 2, b = log23
Then 2log2 3 = 3 is rational.
(C) a = 3 , b = 2
( )
3
2
= 3 is rational.
(D) a = 2, b = 3
Then (2)3 = 8 is rational.
B=
log 3 12 log 3 4
1
2
3
+
−
, C=
−
. Find B + C.
log 3 2 log 9 4 log 27 8
log 36 3 log 108 3
(A) 3
Sol.
(B) 2
(C) 4
(D) 1
(B)
B = log23 + 2log49 – 3log827
= log 2 3 + 2log 22 32 − 3log 23 33
= log23 + 2log23 – 3log23 = 0
15.
Logarithm
Q.15
C=
log 3 12 log 3 4
−
log 36 3 log 108 3
C = log312 log336 – log34.log3108
C = log3(4×3).log3(4×9) – log34.log3(4×27)
C = (log34 + log33)(log34 + log39) – log34(log34 + log327)
C = (log34 + 1)(log34 + 2) – log34.(log34 + 3)
Put log34 = t,
C = (t + 1)(t + 2) – t(t + 3) = t2 + 3t + 2 – t2 – 3t = 2
So, B + C = 0 + 2 = 2
Q.16 A =
Sol.
1
1
1
+
+
1 + logba + logbc 1 + log ca + log cb 1 + log ab + log a c
Where a > 0, a ≠ 1, b > 0, b ≠ 1, c > 0, c ≠ 1, abc ≠ 1, then A is :
1
(A) abc(B)
(C) 1 abc
(C)
1
1
1
+
+
logbb + logba + logbc log cc + log ca + log cb log a a + log ab + log a c
=
(D) 0
1
1
1
+
+
logbabc log cabc log a abc
= logabcb + logabcc + logabca
= logabcabc = 1
Q.17 Let a = log 5, b = log
3
following is/are true?
(A) a > b
Sol. (AC)
a = log35 =
25 and c = log51000, d = log72058 then which of the
17
(B) a < b
(C) c > d (D) c < d
1
2
and b = log1725 = log1752 = 2log175 =
log 5 3
log 5 17
1
1
1
= log 5 3 and = log 5 17 = log 5 17
a
b 2
1
1
From here log 5 17 > log 5 3 ⇒ >
b a
Logarithm
Then
⇒a>b
c = log51000 and d = log7(2058)
∵ 625 < 1000 < 3125
⇒
log5625 < log51000 < log53125
⇒
log554 < c < log555
⇒
4<c<5
343 < 2058 < 2401
16.
⇒
log7343 < log72058 < log72401
⇒
log773 < d < log774
⇒
3<d<4
Hence c > d
Q.18 Establish the trichotomy between:
(i)
Sol.
m = (log25)2 & n = log220
m – n = (log25)2 – (log220)
= (log25)2 – log2(5×4)
= (log25)2 – (log25 + 2log22)
= (log25)2 – log25 – 2
Let log25 = t
then, m – n = t2 – t – 2 = (t –2)(t + 1)
log25 > 2 ⇒ t > 2 hence m – n > 0 ⇒ m > n
Q.
(ii)
Sol.
Suppose α = logπ2
a = logπ2 + log2π and b = 1
1
> 2 and b = 1
α
Then a = α +
So a > b
Q.
If log615 = α and log1218 = β, then find log2524 in terms of α, β.
Sol.
log 3 15
⇒
1 + log 3 5
1 + log 3 2
Also
⇒
log 3 ( 2 × 3 )
=α⇒
log 3 3 + log 3 5
log 3 3 + log 3 2
=α
= α ... (i)
log 3 18
log 3 12
2 + log 3 2
log 3 ( 3 × 5 )
1 + 2log 3 2
= β⇒
log 3 ( 9 × 2 )
log 3 ( 3 × 4 )
= β⇒
log 3 9 + log 3 2
log 3 3 + log 3 4
=β
= β ... (ii)
Now, log 25 24 =
log 3 24 log 3 ( 8 × 3) 1 + 3log 3 2
=
=
log 3 25
log 3 52
2log 3 5
By equation (ii), 2 + log32 = β + 2βlog32
⇒ (log32)(1 – 2β) = β – 2
β−2
⇒ log 3 2 =
1 − 2β
17.
Logarithm
log 3 6
=α ⇒

β−2 
By equation (i), 1 + log35 = α(1 + log32) = α 1 +

 1 − 2β 
 1 − 2β + β − 2 
 −1 − β 
⇒ 1 + log35 = α 
 = α

1 − 2β


 1 − 2β 
−αβ − α − 1 + 2β
1 − 2β
⇒ log 3 5 =
1 + 3log 3 2
=
2log 3 5
Then log 24 25 =
⇒ log 25 =
24
Q.20 If log 12 = a and log
7
Sol.
 2β − 1 − αβ − α 
2

1 − 2β

1 − 2β + 3β − 6
β−5
=
2 2β − 1 − αβ − α
2 2β − 1 − αβ − α
(
(
)
)
24 = b then find the value of log54168 in terms of a, b.
12
Given, a = log 7 12 =
and b = log1224 =
log54168 =
 β−2
1 + 3
 1 − 2β 
log 2 12 log 2 ( 4 × 3) 2 + log 2 3
… (i)
=
=
log 2 7
log 2 7
log 2 7
log 2 24 log 2 ( 8 × 3) 3 + log 2 3
=
=
… (ii)
log 2 12 log 2 ( 4 × 3 ) 2 + log 2 3
log 2 168 log 2 ( 7 × 3 × 8) 3 + log 2 7 + log 2 3
=
=
log 2 54
log 2 ( 27 × 2)
1 + 3log 2 3
… (iii)
From equation (ii), 3 + log23 = 2b + b log23
⇒ (1–b)log23 = 2b – 3
2b − 3
⇒ log 2 3 =
1−b
Replace this in equation (i),
2b − 3
2+
1 − b = log 7
2
a
2 − 2b + 2b − 3
⇒
= log 2 7
a ( 1 − b)
⇒ log 2 7 =
−1
a ( 1 − b)
3+
Logarithm
Then from equation (iii), log 54 168 =
⇒ log 54 168 =
2b − 3
1
−
1−b
a ( 1 − b)
2b
−
3

1 + 3

 1−b 
3a ( 1 − b ) + a ( 2b − 3) − 1
a ( 1 − b ) + 3 ( 2b − 3 ) 
18.
⇒ log 54 168 =
⇒ log 54 168 =
Q.21
3a − 3ab + 2ab − 3a − 1
a [1 − b + 6b − 9]
−1 − ab
(
)
a 5b − 8




If log 7  log 7 7 7 7  = 1 – a log72 and log 15  log 15 15 15 15 15  = 1 – b log152.




Find a + b.
Let


x = log 7  log 7 7 7 7 


1

x = log 7  log 7 7 7 7 
2

(
)
1

x = log 7  log 7 7 + log 7 7 7 
2

1
1

x = log 7   1 + log 7 7 7  
2

2
1
1
1

x = log 7   1 +  log 7 7 + log 7 7   
2
2

2
1
1
1 
x = log 7   1 +  1 +   
2
2 
2
1 3 
1
x = log 7   1 + .  
2 2 
2
7
 1  7 
x = log 7     = log 7
2
4
8
  
x = 1 – log78
x= 1 – 3log72
Compare it with x = 1 – a log72 then a = 3


Similarly, suppose y = log 15  log 15 15 15 15 15 


then
1

y = log 15  log 15 15 15 15 15 
2


1
1

y = log 15   1 + log 15 15 15 15  
2

2
1 
1
1
 
y = log 15   1 +  1 + log 15 15 15   
2
2
 
2 
19.
Logarithm
Sol.
1 
1
1
1   
y = log 15   1 +  1 +  1 +    
2
2
2   
2 
1 
1
1 3  
y = log 15   1 +  1 + .   
2
2 2  
2 
7 
1 
 1 15 
y = log 15   1 +   = log 15  . 
8 
2 8 
2 
4
y = log1515 – log152
y = 1 –4 log152, compare it with y = 1 – b log152
Hence b = 4 and a = 3
So
a+b=4+3
a+b=7
Q.22
If
Sol.
Assume,
log a log b log c
, show that aa.bb.cc = 1.
=
=
b−c
c−a
a −b
log a log b log c
=
=
=k
b−c
c−a
a −b
then
log a = k(b –c) ⇒ a log a = k (ab – ac)
log b = k(c –a) ⇒ b log b = k (bc – ab)
log c = k(a –b) ⇒ c log c = k (ac – bc)
Adding (i), (ii) and (iii), we get
⇒ log(aa.bb.cc) = k[ab – ac + bc – ab + ac – bc]
⇒ log(aa.bb.cc) = 0 (change it into exponential form)
⇒ aa.bb.cc = 1
Q.23
If a, b, c are positive real numbers other than unity such that
a (b + c − a ) b ( c + a − b ) c ( a + b − c )
=
=
, prove that abba = bccb = caac.
log a
log b
log c
Sol.
Assume,
a (b + c − a )
log a
⇒ log a =
log b =
log c =
Logarithm
… (i)
… (ii)
… (iii)
=
b (c + a − b)
log b
a (b + c − a )
k
b (c + a − b)
k
c (a + b − c )
⇒ b log a =
k
c (a + b − c )
log c
…(i)
…(ii)
…(iii)
ab (b + c − a )
k
=
, a log b =
=k
ab ( c + a − b )
k
, log c =
c (a + b − c )
k
20.
⇒ log ab =
log ba =
ab (b + c − a )
k
…(iv)
ab ( c + a − b )
…(v)
k
Add equations (iv) & (v)
2abc
ab
log (ab.ba) =
[b + c – a + c + a –b] =
k
k
Consider the base as ‘e’,
2abc
we get, abba = e k
… (A)
Now multiply (ii) by c and (iii) by b, we get
bc ( c + a − b )
c log b =
… (vi)
k
and b log c =
bc ( a + b − c )
…(vii)
k
Add (vi) and (vii), we get
bc
2abc
log (bc .cb ) =
[c +a – b + a + b – c] =
k
k
Change this into exponential form, we get
bc .cb = e
2abc
k
… (B)
2abc
k
Similarly we get, ca .ac = e
… (C)
From equations (A), (B) and (C), we get
ab.ba = bc.cb = ca.ac
Logarithmic Equation
Find ‘x’ in following :
(i)
x2 + 7log 7 x − 2 = 0
Sol. Using a
= N , we get
x + x – 2 = 0 and x > 0
⇒ x2 + 2x – x – 2 = 0
⇒ x(x + 2) – 1(x + 2) = 0
⇒ (x + 2) (x – 1) = 0
⇒ either x = 1 or x = –2
Since x > 0, so we get x = 1
log aN
2
21.
Logarithm
Q.1
( )
Q.
(ii)
Sol.
Using property alogaN = N
We get x2 –3x – 4 = 0
⇒
x2– 4x + x – 4 = 0
⇒
x(x –4) + 1(x –4) = 0
⇒
(x –4)(x + 1) = 0
⇒
x = 4 and x = –1
log 2 x2
2
− 3x − 4 = 0
log 2 ( 9 − 2x )
Q.
(iii)
Sol.
log2(9–2x) = 3 – x
⇒
9 – 2x = 23–x
3−x
=1
(change it into exponential form)
⇒
⇒
⇒
⇒
⇒
⇒
⇒
8
2x
Put 2x = t, we get
8
9–t=
t
2
9t – t = 8
t2 – 9t + 8 = 0
t2 – 8t –t + 8 = 0
t(t –8) –1(t –8) = 0
t = 8, t = 1
2x = 23 and 2x = 1
x = 3 and x = 0, but x ≠ 3, hence x = 0
Q.
(iv)
( x + 1)
Sol.
By taking logarithm both sides with base 10, we get
⇒
⇒
9 – 2x = 23.2–x =
log 10 ( x + 1)
= 100 ( x + 1)
log 10 ( x + 1)
log 10 ( x + 1)
= log 10 100 ( x + 1)
log10(x + 1).log10(x + 1) = log10100 + log10(x + 1)
Consider, log10(x + 1) = t, we get
t2 = 2 + t
⇒ t2 – t – 2 = 0
⇒ t2 – 2t + t – 2 = 0
⇒ t(t –2) + 1(t –2) = 0
⇒ (t –2)(t + 1) = 0
⇒ t = 2, t = –1
log10(x + 1) = 2 and log10(x +1) = –1
1
⇒ x + 1 = 100 and (x + 1) =
10
Logarithm
⇒
22.
⇒ x = 100 – 1 = 99 and x =
1
−1
10
−9
10
9
⇒ x = 99 and x = −
10
⇒ x = 99 and x =
Q.
(v)
Sol.
logx–122 = 1 + log2(x–1)
2
⇒
= 1 + log 2 ( x − 1)
log 2 ( x − 1)
logx–14 = 1 + log2(x –1)
Consider log2(x –1) = t, we get
⇒
⇒
2 = t + t2
t2 + t – 2 = 0
t2 +2t – t –2 = 0
t(t + 2) –1(t + 2) = 0
(t + 2)(t –1) = 0
t = –2, t = 1
log2(x –1) = –2 and log2(x –1) = 1
(x–1) = 2–2 and (x –1) = 21
1
x=1+
and x = 1 + 2 = 3
4
5
x = and x = 3
4
1
1
1

log + log x + log 5

3
2
3

Q.
(vi) 1 – log 5 =
Sol.
Base is given as 10
log 10 – log 5 =
⇒
⇒
⇒
⇒
⇒
1
1
1

log + log x + log 5

3
2
3

10 1
=  −log 2 + log x + log 51/3 
5
3
3 log 2 + log 2 = log x + log 51/3
4 log 2 – log 51/3 = log x
16
log x = log 1/3
5
16
x = 1/3
5
log
23.
Logarithm
⇒
⇒
⇒
⇒
⇒
⇒
⇒
⇒
2
=1+t
t
Q.
(vii) 3log3x + x(log3x ) = 162
2
Sol. ( 3 )
log 3 x log 3 x
+ xlog3x − 162 = 0
⇒ xlog3x + xlog3x − 162 = 0
⇒ 2xlog3x = 162
⇒ xlog3x = 81
Take logarithm on both sides with base 3, we get
log3x.log3x = log381 = 4
⇒ (log3x)2 = 4
⇒ log3x = ±2
⇒ x = 3±2
⇒ x = 9, 1
9
Q.
(viii) 51+log 4x + 5(log1/ 4x)−1 =
Sol.
51+log 4x + 5−1−log 4x =
26
5
26
5
Put 51+log 4x = t , we get
1 26
t+ =
t
5
26t
⇒
t2 + 1 =
5
2
⇒
5t + 5 = 26t
⇒
5t2 – 26t + 5 = 0
⇒
5t2 – 25t – t + 5 = 0
⇒
5t(t –5) – 1 (t –5) = 0
⇒
(t –5)(5t –1) = 0
⇒
t = 5 or t = 1/5
⇒
⇒
⇒
⇒
Logarithm
⇒
1
= 5−1
5
1 + log4x = 1 or 1 + log4x = –1
log4x = 0 or log4x = –2
x = 1 or x = 4–2
1
x = 1 or x =
16
5
1+log 4 x
= 51 or 5
1+log 4 x
=
24.
Q.
(ix) log4(2log3(1 + log2(1 + 3log2x))) =
1
2
it into exponential form,
Sol. Change
2log (1 + log (1 + 3log x)) = 4 = 2
3
2
2
1/2
Again, 1 + log2(1 + 3log2x) = 3
log2(1 + 3log2x) = 2
Again, 1 + 3log2x = 4
⇒ 3log2x = 3
⇒ log2x = 1
⇒x=2
Q.
log 5 ( 51/x + 125 ) = log 5 6 + 1 +
(x)
Sol. log (5
5
1/x
1
2x
1 

+ 125) = log56 +  1 +
 log55
2x 

⇒ log5(5
1/x
1 

 1+ 
 2x 
+ 125) = log56 + log 5 5
⇒ log5(5 + 125) = log5(6×5×51/2x)
⇒ 51/x + 125 = 30×51/2x
Put 51/2x = t, then t2 + 125 = 30t
⇒ t2 – 30t + 125 = 0
⇒ t2 – 25t – 5t + 125 = 0
⇒ t(t –25) – 5(t –25) = 0
⇒ (t–25)(t –5) = 0
Either t = 25 or t = 5
⇒ 51/2x = 52 or 51/2x = 51
1
1
= 2 or
=1
⇒
2x
2x
1
1
⇒ x = or x =
4
2
1/x
Sol.
(xi)
x2+log x 4 = x2 + 27
x2 .xlog x 4 = x2 + 27
⇒ x2.4 = x2 + 27
⇒ 3x2 = 27
⇒ x2 = 9
⇒ x = +3 or x = –3
As x > 0 and x ≠ 1, so we get x = 3
25.
Logarithm
Q.
Q.
(xii) a2log2x = 5 + 4xlog2a
Sol. (a
)
log 2 x 2
= 5 + 4 ( alog2x ) (Using property alogcb = blogca )
Put alog2x = t , we get
t2 = 5 + 4t
⇒ t2 – 4t – 5 = 0
⇒ t2 – 5t + t – 5 = 0
⇒ t(t –5) + 1(t–5) = 0
⇒ (t–5)(t + 1) = 0
⇒ t = 5 or t = –1
⇒ alog2x = 5 or alog2x = −1 (not possible)
⇒ log5a.log2x = log55
⇒ log2x.log5a = 1
1
⇒ log 2 x =
log 5a
⇒ log2x = loga5
⇒ x = 2log 5 or 5log
a
a
2
Common logarithm and Natural logarithm:
log10N is referred as a common logarithm.
logeN is called as a natural logarithm. It is also
written as lnN.
Note:
e is an irrational quantity lying between 2.7 to
2.8. Also elnx = x
Characteristics and Mantissa
Logarithm
Real number = Integer + Fraction
logaN = Characteristics + Mantissa
R=I+f
where
R is a real number
I is an integer
f is a fraction i.e. 0 ≤ f < 1
Here, characteristics is
mantissa is fraction, so
0 ≤ mantissa < 1
an
integer
Ex. (i) 3.4 = 3 + 0.4
R = 3.4, I = 3 and f = 0.4
(ii)
5.1 = 5 + 0.1
R = 5.1, I = 5 and f = 0.1
(iii)
–1.2 = –1 –0.2
–1.2 = –1 – 1 + 1 –0.2
–1.2 = –2 + 0.8
R = –1.2, I = –2, f = 0.8
(iv)
–3.8 = –3 – 0.8
–3.8 = –3 –1 + 1 –0.8
–3.8 = –4 + 0.2
R = 3.8, I = –4, f = 0.2
(v)
–6.2 = –7 + 0.8
R = –6.2, I = –7, f = 0.8
and
26.
Note:
(i) Characteristic can be zero, positive and negative.
(ii) Mantissa is always non-negative and less than 1.
Finding Number of digits
Number of digits = Characteristics + 1
Observation:
Range
Taking log
Characteristics
Number of digits
1 ≤ N < 10
0 ≤ log10N < 1
0
1
10 ≤ N < 100
1 ≤ log10N < 2
1
2
100 ≤ N < 1000
2 ≤ log10N < 3
2
3
By observation, we can say that
Number of digits = Characteristics + 1
Remember the following values
log102 = 0.3010
log103 = 0.4771
log105 = 0.6989
log107 = 0.8450
Q.1
Find number of digits in
(i) N = 650
log10N = log10650 = 50 log106
27.
Logarithm
Sol. Taking logarithm on both sides, we get
= 50[log102 + log103]
= 50[0.3010 + 0.4771]
= 50[0.7781] = 38.905
log10650 = 38 + 0.905
Characteristics = 38, So Number of digits = 39
Q.
(ii)
N = 525
Taking logarithm on both sides, we get
10
N = log10525 = 25log10
= 25[log1010 – log102]
10
2
log10N = 25[1–0.3010]
= 25[0.699]
log10525 = 17.475 = 17 + 0.475
Here characteristics = 17
Hence, Number of digits = 18
Sol. log
Number of zeroes after decimal before a significant digit starts:
Number of zeroes = |Characteristics + 1|
Observation:
Range
Taking log
Characteristics
Number of zeroes
0.1 ≤ N < 1
–1 ≤ log10N < 0
–1
0
0.01 ≤ N < 0.1
–2 ≤ log10N < –1
–2
1
0.001 ≤ N < 0.01
–3 ≤ log10N < –2
–3
2
Logarithm
Hence by observation we can say that
Number of zeros = |Characteristics + 1|
28.
Q.2
Find number of zeroes after decimal point before a significant digit starts in
(i) N = 3–50
Sol. Taking logarithm on both sides, we get
⇒
⇒
Q.
(ii)
log10N = log103–50 = (–50)log103
log10N = (–50) × (0.4771) = –23.855
log10N = –23 – 0.855 + 1 – 1 = –24 + 0.145
Here characteristics = –24
Hence Number of zeros = |–24 +1| = |–23| = 23
9
N= 
8
−100
Sol. Taking logarithm on both sides with base ‘10’
9
log10N = log 10  
8
−100
9
= −100log 10   = –100[2log3 – 3log2]
8
⇒ log10N = –100[2 × 0.4771 – 3 × 0.3010]
⇒ log10N = –100[0.9542 – 0.9030]
⇒ log10N = –5.12
Characteristics = –6
Number of zeros = |–6 + 1| = |–5| = 5
Q.3
Find ‘N’ for which characteristics in 2 and base is 10
(A) N ∈ [1000, 10000)
(B) N ∈ [10, 100)
(C) N ∈ [100, 1000)
(D) N ∈ [1, 10)
Sol. (C)
According to given condition,
⇒
2 ≤ log10N < 3
⇒
100 ≤ N < 1000
⇒
N ∈ [100, 1000)
Q.4
Find the number of integral values of N, for which characteristic is 5 under
base 3.
to given condition,
Sol. According
5 ≤ log N < 6
3
⇒
35 ≤ N < 36
⇒
243 ≤ N < 729
Number of integral values of N = 729 – 243 = 486
Find ‘N’ for which characteristics is –5 under base 3.
to given statement,
Sol. According
–5 ≤ log N < –4
⇒
3
3–5 ≤ N < 3–4
29.
Logarithm
Q.5
1
1
≤N< 4
5
3
3
1
1
⇒
≤N<
243
81
⇒
1 
 1
, 
⇒ N∈ 
243
81


Significance of Modulus function in Logarithmic Equation:
Q.1
Solve : log4x2 =2
Sol. xx
=42
= 16
x = ±4
Wrong method:
log4x2 = 2
⇒ 2log4x = 2
⇒ log4x = 1
⇒ x = 41
⇒x=4
Incomplete solution
That’s why
log(x2) = 2log |x|
Now it can be solved as
log4x2 = 2 ⇒
2log4|x| = 2
⇒
log4|x| = 1
⇒
|x| = 41
⇒
x=±4
2
2
Note:
(i) logx2 = 2 log|x|
(ii) logx2m = 2m log|x|, m ∈ N
(iii) logx2m+1 = (2m +1) logx, m ∈ N
(iv) x2 = x
y
Absolute value function/modulus function
y = –x
Logarithm
Y
x ; x 0
x x 0 ; x 0
x ; x0
y=x
2
(0, 0)
x
30.
Q.2
Solve : |x–5| = 10. Find ‘x’
Sol.
Q.3
Sol.
x – 5 = ± 10
⇒ x – 5 = 10 or x – 5 = –10
⇒ x = 15 or x = –5
Find ‘x’ if |3x – 2| + x = 11
2
3
Then, –(3x –2) + x = 11
⇒
–3x + 2 + x = 11
⇒
–2x = 9
⇒
x = –9/2
When x <
Case-(i)
–
–∞
+
∞
2
3
2
3
(3x –2) + x = 11
⇒
4x = 13
⇒
x = 13/4
Hence, x = –9/2 and x = 13/4 are the solution.
Case-(ii)
Sol.
Find ‘x’ if |x – 3| +2|x + 1| = 4
∞
–∞
–1
3
Case-(i)
⇒
⇒
⇒
Case-(ii)
⇒
⇒
Case-(iii)
⇒
⇒
When x <–1, then
–(x –3) –2(x + 1) = 4
–x + 3 – 2x – 2 = 4
–3x = 3
x = –1, but x < –1 hence no solution.
When –1 ≤ x < 3, then
–(x –3) + 2(x + 1) = 4
–x + 3 + 2x + 2 = 4
x = –1 is a solution.
When x ≥ 3, then
(x –3) +2(x +1) = 4
x –3 + 2x + 2 = 4
3x = 5
5
x=
⇒
but x ≥ 3
3
Hence no solution
So, x = –1 is the only solution.
31.
Logarithm
Q.4
When x ≥
Q.5
Find ‘x’
Sol.
2log 8 2x log 8 x 1 (i)
2log8(2x) + log8(x2 + 1 – 2x) =
2
4
3
4
3
4
2log 8 2x 2log
x1 8
3
2
log 8 2x x 1 3
(Change it into exponential form)
2
2x x 1 8 3 4
x x1 2
Case (i): When x < 1, then
–x (x – 1) = 2
⇒ x2 – x + 2 = 0
It’s discriminant D = (–1)2 – 4 × 2 = 1 – 8 = –7 < 0
Hence, no real roots exist, so no solution.
Case (ii): When x ≥ 1, then
x (x –1) = 2
⇒ x2 – x = 2
⇒ x2 – x – 2 = 0
⇒ (x –2)(x + 1) = 0
⇒ x = – 1 or x = 2, but x ≥ 1
So, x = 2 is the only solution.
Q.
(ii)
Sol.
2log 3 ( x − 2) + 2log 3 x − 4 = 0
2log 3 ( x − 2 ) + log 3 ( x − 4 ) = 0
2
⇒ log 3 ( x − 2 ) x − 4 = 0 ⇒ (x – 2) |x – 4| = 1
Case (i): When x < 4, then
Logarithm
− ( x − 2 )( x − 4 ) = 1
⇒ x2 − 6x + 8 = −1
⇒ x2 − 6x + 9 = 0
⇒ (x – 3)2 = 0 ⇒ x = 3
(Change it into exponential form)
–∞
∞
4
32.
Case (ii): When x ≥ 4
( x − 2)( x − 4 ) = 1
⇒ x2 − 6x + 8 = 1
⇒ x2 − 6x + 7 = 0
⇒x=
6± 8
2
⇒ x = 3 + 2, x = 3 − 2
but x ≥ 4
So, x = 3 + 2
Hence, x = 3 or x = 3 + 2 is the solution.
(
)
(
)
(iii) log 4 x2 − 1 − log 4 x − 1
Sol. log ( x
4
2
⇒ log 4
⇒
(x
2
)
2
= log 4
(4 − x)
2
− 1 − log 4 ( x − 1) = log 4 4 − x
(x
2
2
) = log
−1
( x − 1)
2
4
4 − x = log 4 x − 4
) = x−4
−1
( x − 1)
( x − 1)( x + 1) = x − 4
⇒
( x − 1)
( x + 1) = x − 4
⇒
( x − 1)
2
2
Case (i):
Case(ii):
–∞
When x < 4
4
∞
x+1
= − ( x − 4)
x−1
⇒ −x − 1 = ( x − 1)( x − 4 )
⇒ –x – 1 = x2 – 5x + 4
⇒ x2 – 4x + 5 = 0, Since D < 0
Hence no real roots.
When x ≥ 4
x+1
= x−4
x−1
(
)
33.
Logarithm
Q.
(
(
) (
)
)(
)
⇒ x+1 = x−4 x−1
⇒ x + 1 = x2 − 5x + 4
⇒ x − 6x + 3 = 0
⇒ x =3± 6
⇒ x = 3 + 6,3 − 6
Since x ≥ 4 so x = 3 + 6
Q.
(iv)
x−2
10x2 − 1
=x−2
3x
Sol. Comparing the powers,
10x2 − 1 = 3x
⇒ 10x2 − 3x − 1 = 0
⇒ 10x2 − 5x + 2x − 1 = 0
5x ( 2x − 1) + ( 2x − 1) = 0
⇒ ( 2x − 1)( 5x + 1) = 0
⇒x=
1
1 or
x=−
5
2
When base is equal to 1, we get
|x –2| = 1
⇒x–2=±1
⇒ x = 3 and x = 1
Q.
(v)
| x − 3 |3x
Sol. | x − 3 |
Logarithm
3x2 − 10x + 3
2
− 10x + 3
=1
0
= x − 3 (Comparing the powers)
2
⇒ 3x – 10x + 3 = 0
⇒ 3x2 – 9x – x + 3 = 0
⇒ 3x(x –3) –1(x –3) = 0
⇒ (x – 3)(3x – 1) = 0
1
⇒x= ,3
3
When base is equal to 1,
|x – 3| = 1
⇒x–3=±1
⇒ x = 4, 2
When x = 3, base is 0, so we reject this.
1
So x = 4, 2,
is the solution.
3
34.
Graphs of logarithm:
f ( x ) = log a x, x > 0, a ≠ 1, a > 0
Case-1:
a>1
Ex.: f x = log 2 x = y
y
()
x=
x=
x
x
x
x
x
=
=
=
=
=
1 ,
2
1
,
4
1,
2,
4,
8,
16,
y = logax, when a > 0, a 0
y = –1
y = –2
y
y
y
y
y
=
=
=
=
=
x
(1, 0)
0
1
2
3
4
y
Case-2:
0<a<1
Ex.: f (x ) = log 1 x = y
x
x
x
x
=
=
=
=
2
2,
4,
8,
16,
y =f(x) = logax
where 0 < a < 1
(1, 0)
y
y
y
y
=
=
=
=
–
–
–
–
x
1
2
3
4
Note:
1. W
henever the number and base are on
the same side of unity then logarithm of
that number to the same base is positive.
(ii)
y = f ( x ) = log a x > 0 ⇒ x, a are on the same side of unity.
y = f ( x ) = log a x < 0 ⇒ x, a are on the opposite side of unity.
35.
Logarithm
(i)
2. W
henever the number and base are on
the opposite side of unity then logarithm
of that number to the base is negative.
Q.1
17


2
2
Solve log 4 log 3 x + log 1  log 1 y  = 0 and x + y =
.
4


4 
3 
Sol.


log 4 (log 3 x ) + log 1  log 1 y  = 0


4 
3 
(
x2 + y 2 =
)
17
4
.... (i)
....(ii)


log 4 (log 3 x ) − log 4  log 1 y  = 0


3 



⇒ log 4 (log 3 x ) = log 4  log 1 y 


3 

log 3 x = log 1 y
3
log 3 x = − log 3 y
log 3 x = log 3
x=
1
y
1
y
Equation (ii), x2 +
t+
1 17
=
t
4
⇒
t 2 + 1 17
=
t
4
1
17
=
, put x2 = t
2
4
x
⇒ 4t 2 + 4 = 17t
⇒ 4t 2 − 17t + 4 = 0
⇒ t = 4,
1
4
⇒ x2 = 4,
1
4
Logarithm
⇒ x = ±2, ±
1
1
but x > 0 ⇒ x = 2 and y =
2
2
36.
Graphs of Exponential
f ( x ) = ax
Case-1: a > 1
a > 0,
a≠1
y
f(x) = ax
a>1
(0, 1)
x
( )
x
Ex.: y = f x = 2
x = 1,
x = 2,
x = 3,
Case-2:
y=2
y=4
y = 8
Increasing
0<a<1
y
y = ax = f(x)
0<a<1
(0, 1)
x
Ex.:
 1
f (x) =  
2
x
1
2
x = 1,
y=
x = 2,
y=
1
4
x = 3,
y=
1
8
Decreasing
()
x → −∞ ⇒ f (x ) → ∞
x→∞ ⇒f x →0
37.
Logarithm
Logarithmic Inequalities
1.
f ( x ) = log a x
(i) When a > 1 and x1 > x2 ⇔ log a x 1 > log a x2
(ii) When 0 < a < 1 and x1 > x2 ⇔ log a x 1 < log a x2
2.
f ( x ) = ax
(i) When a > 1 and x1 > x2 ⇔ a
x1
>a
(ii) When 0 < a < 1 and x1 > x2 ⇔ a
Q.1
<a
x2
Find ‘x’
log 1 2x + 1 > 0
(i)
2
Sol.
x1
x2
(
)
log 1 ( 2x + 1) > log 1 1
2
⇒ ( 2x + 1) < 1
2
⇒x<0
and
( 2x + 1) > 0
⇒x>−
1
2
Logarithm
 1 
By Intersection, x ∈  − , 0 
 2 
Q.
(ii)
Sol.
log 2 ( x + 2) > log 2 1
log 2 ( x + 2) > 0
⇒ (x + 2) > 1
⇒x>–1
and x + 2 > 0
⇒x>–2
By intersection, we get x > – 1 or x ∈ (–1, ∞).
38.
Q.
Sol.
(iii)
23−6x > 1
23−6x > 20
⇒ 3 – 6x > 0
⇒ 6x < 3
3
6
1
⇒x<
2
⇒x<

1
x ∈  –∞, 
2

Q.
Sol.
 1
 
5
(iv)
2x + 1
> 125
− ( 2x + 1)
5
> 53
⇒ −2x − 1 > 3
⇒ −2x > 4
⇒ x < −2
⇒ x ∈ ( −∞, −2 )
Q.
(v)
Sol.
log 0.3 x − 2 > log 0.3 1
log 0.3 x − 2 > 0
⇒ |x – 2| < 1
⇒–1<x–2<1
⇒1<x<3
and |x –2| = 1 ⇒ x – 2 = ± 1 ⇒ x = 3 or 1
and |x – 2| ≠ 0 ⇒ x ≠ 2
By Intersection, x ∈ 1, 2 ∪ 2, 3 or x ∈ 1, 3 – 2
( ) ( )
Q.
(vi)
Sol.
log 8 x2 − 4x + 3 ≤ log 8 8
(
( ) {}
)
log 8 x2 − 4x + 3 ≤ 1
(
)
⇒ x2 − 4x + 3 ≤ 8 and x2 − 4x + 3 > 0
⇒ x2 − 5x + x − 5 ≤ 0 and x2 − 3x − x + 3 > 0
39.
Logarithm
2
⇒ x2 − 4x − 5 ≤ 0 and x − 4x + 3 > 0
⇒ x (x-5) + 1 (x-5) ≤ 0 and x ( x − 3) − 1 ( x − 3) > 0
⇒ ( x − 5 )( x + 1) ≤ 0 and ( x − 3 )( x − 1) > 0
(
) ( )
x ∈  −1, 1) ∪ ( 3, 5
⇒ x ∈  −1, 5 and x ∈ −∞, 1 ∪ 3, ∞
By intersection,
Q.

 x2 + x  
log
log

(vii)
  < 0
1 
6 
 x + 4 
2 
Sol. (i)
 x2 + x   1 
log 6 
> 
 x + 4  2
0
x2 + x
>6
x+4
(ii)
 x2 + x 
log 6 
>0
 x+4 
x2 + x
>1
x+4
x2 + x
>0
x+4
By intersection of (i), (ii) and (iii) we can clearly state that
x2 + x
>6
x+4
(iii)
⇒
x2 + x
−6 >0
x+4
⇒
x2 + x − 6x − 24
>0
x+4
⇒
x2 − 5x − 24
>0
( x + 4)
( x − 8)( x + 3) > 0
( x + 4)
⇒ x ∈ ( −4, −3 ) ∪ ( 8, ∞ )
Logarithm
⇒
40.
Q.

 3x +6  
log 1  log 2 
 
2
 x +2  
3
( )
(viii) 0.3
Sol. (0.3)
>1
3x + 6
0
 3x + 6 
> ( 0.3 ) and log 2  2
 > 0 and x2 + 2 > 0
x +2

 3x + 6  
3x + 6
3x + 6
>0
> 1 and 2
log 1  log 2  2
  < 0 and 2
x
+
2
x
2
+
x
2
+




3

 3x +6  
log 1  log 2 

x2 + 2  


3
So,
 3x + 6 
⇒ log 2  2
>1
x +2
3x + 6
>2
x2 + 2
By intersection, we get
3x + 6
>2
x2 + 2
⇒ 3x + 6 > 2(x2 + 2)
⇒ 2x2 + 4 < 3x + 6
⇒ 2x2 – 3x – 2 < 0
⇒ 2x2 – 4x + x – 2 < 0
⇒ 2x(x –2) + 1(x – 2) < 0
⇒ (x –2)(2x + 1) < 0
 1 
⇒ x ∈  − , 2
 2 
⇒
3
Q.2
  1 x

  1 x

Find the solution of the equation, 2log 9  2   − 1  = log 27    − 4  . Also
 2

 4 





state whether the solution is rational or irrational.
Sol.
 1
 1
Suppose   = t so   = t 2
2
4
x
x
2
3
log 3 ( 2t − 1) = log 3 t 2 − 4
2
3
(
2t – 1 = t2 – 4
t2 – 2t – 3=0
(t – 3) (t + 1) = 0
t = 3, t = – 1
x
x
 1
 1
⇒   = 3 or   = – 1
2
2
(Not possible)
41.
Logarithm
⇒
⇒
⇒
⇒
)
⇒ 2− x = 3
⇒ – x log2 = log3
⇒x= −
log 3
log 2
( )
⇒ x = − log 2 3
Q.3
which is irrational.
If the product of all solutions of equation
expressed in the lowest form as
( 2020) x =
2021
( 2020)
log x 2021
can be
m
. (m, n ∈ I) then find the value of (m + n).
n
 2020x 

 = log x ( 2021)·log ( 2021) ( 2020 )
 2021 
⇒ log 2021 2020 + log 2021x − 1 = log x 2021·log 2021 2020
Sol. log
2021
Suppose log2021x = t then log x 2021 =
Then, log 2021 2020 + t − 1 =
1
t
1
log 2021 2020
t
t ·log 2021 2020 + t 2 − t − log 2021 2020 = 0
⇒
⇒
⇒
⇒
(t – 1) log2021(2020) + t(t – 1) = 0
(t – 1) (log20212020 + t) = 0
t = 1 or t = –log20212020
log2021x = 1 or log2021x = log2021(2020)–1
⇒ x = 2021
or
x=
1
2020
Product of solutions
m 2021
=
n 2020
m + n = 2021 + 2020 = 4041
Q.4
The sum of the integral values(s) of a ∈ [–9, 9] so that equation
(x–2)log133 + log13(3x – 7a) = log132 + 2log13 a has integral solution.
Sol. log
13
(
)
(
3x −2 · 3x − 7a = log 13 2·a 2
(
)
)
⇒ 3x −2 3x − 7a = 2a 2
Put 3 = t, we get
t
t − 7a = 2a 2
9
Logarithm
x
(
)
⇒ t 2 − 7at = 18a 2
42.
⇒ t 2 − 9at + 2at − 18a 2 = 0
⇒ (t – 9a) (t + 2a) = 0
⇒ t = 9a or
t = – 2a
⇒ 3x = 9a or
3x = – 2a
30, 31, 32, 33, ... = – 2a
or
1, 3, 9, 27, ... = –2a
or
30, 31, 32, ... = 9a
1, 3, 9, ... = 9a
1 3 9
a= ,
,
,…
No integral value of a
9 9 9
a=
1 1
,
, 1, 3, 9, …
9 3
a = 1, 3, 9
Sum = 1 + 3 + 9 = 13
Q.5
or i = 1 to 6, let loga(logb(logcxi))=0, where a, b and c represent every possible
F
different arrangement of 2, 4 and 8. The product x1x2x3x4x5x6 can be expressed
in the form 2N. The value of N, is
(A) 20
(B) 28
(C) 33 (D) 50
Sol. (D)
43.
Logarithm
loga(logb(logcxi)) = 0
0
⇒ logb(logcxi) = a = 1
⇒ logcxi = b
⇒ xi = cb
a = 2
a = 4
a=8
x = 48, 84 x = 28, 82 x = 24, 42
x1x2x3x4x5x6 = 82 · 48 · 84 · 28 · 42 · 24 = 216 · 212 · 28 · 26 · 24 · 24 = 250
N = 50
44.