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1. Logarithm
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 = a
x , a > 0, a ≠ 1, N > 0
a → base, x → exponent
Illustrations:
Q1. Express 16 using exponential form with base 2, base 4 and base 16.
Sol. (i) N = 16, a 1
= 2, Suppose x 1
= power
2 x 1 = 16 ⇒ 42 = 16
(ii) a 2 = 4, power = x 2
4 x 2 = 16 ⇒ 42 = 16
(iii) a 3 = 16, power = x 3
x 3 1 16 = 16 ⇒ 16 = 16
Q2. Express 81 using exponential form with base 3, base 9, and base 81.
Sol. (i) N = 81, a 1 = 3, x 1 = Power
3 x 1 = 81 ⇒ 43 = 81
(ii) a 2 = 9, x 2 = Power
x 2 2 9 = 81 ⇒ 9 = 81
(iii) a 3
= 81, x 3
= Power
81 x 3 = 81 ⇒ 811 = 81
Q3. Express 16 using exponential form with base ‘3’.
Sol. It means, 3
x = 16
We know that, 3^2 = 9 and 3^3 = 27
⇒ 2 < x < 3 ⇒ x ∈ (2, 3)
Note: (i) What if a = 1.
Express 16 in exponential form with base a = 1
1
x = 16 ⇒ there does not exist any ‘x’.
For which 1x^ = 16
So we can not take base a = 1.
(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.
The types of ‘x’ can be find out
using logarithm.
Example:
Suppose a = –2 and x=
− then
1 2 2
− − does not give real number.
Logarithm 2.
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 ⇔ a
x = N, N > 0, a > 0, a ≠ 1
a → base, x → exponent, N → number
Illustrations:
Q1. Find values of following:
(i) log 10
Sol. Suppose log 10
1000 = x ⇔ 10 x^ = 1000
⇒ 10 x^ = 10^3 (by comparing)
⇒ x = 3
(ii) log 2
Sol. Suppose log 2
32 = y ⇔ 2 y^ = 32
⇒ 2 y^ = 2^5 (by comparing)
⇒ y = 5
(iii) log 5
Sol. Suppose log 5
625 = x ⇔ 5 x^ = 625
⇒ 5 x^ = 5^4 (by comparing)
⇒ x = 4
(iv) 2
log 64
Sol. Suppose 2
log 64 = y ⇔^ ( )
y 2 = 64
y 2 6 2 = 2
y 6 2
= (by comparing)
⇒ y = 12
(v) 2 3
log 1728
Sol. Suppose ( )
x 3 2 3
log 1728 = x ⇔ 2 3 = 1728 = 12
( ) ( )
x 23 2 3 2 3
( ) ( )
x 6 ⇒ 2 3 = 2 3 (by comparing)
⇒ x = 6
(vi) log 1632
Sol. Suppose log 16
32 = y ⇔ 16 y^ = 32
⇒ (
4 )
y = 2
5
⇒ 2 4y^ = 2^5
⇒ 4y = 5 (by comparing)
⇒ y =
Logarithm 4.
Q4. Let a = log1/2 16
b = log 3
(tan30°)
c = log 2 − 3 ( 2 + (^3) )
d = log 2 (log 2 4)
then a + b + c + d is
(A)
− (B)
− (C)
− (D) –
Sol. B
a = log1/2 16
a 1 16 2
⇒ 2 –a^ = 2^4
⇒ a = – 4
b = log 3 (tan30°)
1 b (^12) 3 tan30 3 3
− = ° = =
b 2
c = log 2 − 3 ( 2 + (^3) )
( )
( ) ( )
( )
c 2 3 2 3 2 3 2 3
( ) (^ )^ ( )
c 1 2 3 4 3 2 3
− ⇒ − = − −
c = –
d = log 2
(log 2
we know that log 2
so, d= log 22 ⇒ d = 1
then
a b c d 4 1 1 2 2
Important Deduction:
(i) logNN = 1
(ii) logN1 = 0
(iii) log1/N N = –
5. Logarithm
(i) Proof : Suppose log N
N = x
x ⇔ N = N^1 ⇒ x = 1
(ii) Proof : Suppose log N
1 = y
y 0 ⇔ N = 1 = N ⇒ y = 0
(iii) Proof : Suppose 1 N
log N =z
z 1
N
⇔
z 1 N N N
− = ⇒ = ⇒ z = –
Q5. Find values of following
(i) logsin30°cos60°
Sol. First simplify this, we get (^1)
2
log 2
so (^1)
2
log
(ii) log 4/
Sol. Let N = 1.3 = 1.3333….
N = 1.3333… ...(i)
10 N = 13.33333....... ...(ii)
Subtracting (i) from (ii), we get
–12 4 N –9 3
We get (^4)
3
log 1 3
(iii) log 5 5 5 5...
Sol.
N
N = 5 5 5...
N = 5 N
N
2 = 5N
⇒ N^2 –5N = 0
⇒ N(N–5) = 0
⇒ N = 0 or N = 5 but N = 0 is not possible
We get log 5
(iv) log 2 (sin
2 x + cos
2 x)
Sol. We know that sin^2 x + cos^2 x = 1
So we get , log 2 1 = 0
(v) log(tan1°).log(tan2°).log(tan3°)…log(tan89°)
Sol. log(tan1°). log(tan2°)… log(tan45°)…log(tan89°)
= log(tan1°). log(tan2°)… log1…log(tan89°)
= 0
we know that logNN =
we know that logNN =
(^) log N
7. Logarithm
Q6. If log 2 (log 2 (log 3 x)) = 0 = log 2 (log 3 (log 2 y)) then find x + y.
Sol. log 2
(log 2
(log 3
x) = 0 ⇔ log 2
(log 3
x) = 2^0 = 1 (Change it into exponential form)
⇔ log 3 x = 2
1 = 2
⇔ x = 3^2 = 9
x = 9
Similarly, log 2 (log 3 (log 2 y) = 0 ⇔ log 3 (log 2 y) = 2
0 = 1 (Change it into exponential form)
⇔ log 2
y = 3^1 = 3
⇔ y = 2
3 = 8
Then x + y = 9 + 8 = 17
Fundamental Identify:
alog N a =N
Proof:
Let logaN = x then it changes into
a
x = N ⇔ logaN = x
So we get
log Na a =N
Q Find the value of following: 3
log 10 3
Sol. Using Identify,
log N a =N
We get 3
log 10 3 = 10
Properties of logarithm:
- If m, n are positive real numbers, a > 0, a ≠ 1 then
(i) (^) log amn = logam + logan
Proof:
Suppose logam = x and logan =y
ax^ = m and ay^ = n
then mn = a
x .a
y = a
x + y
again change it into logarithmic form, x + y = log a
mn
we get logamn = logam + logan
Q1. Solve : log 10 2 + log 105
Sol. log 10
2 + log 10
= log 10 (2×5) (using P(1))
= log 1010
= 1
Logarithm 8.
Note:
General version:
Suppose a > 0, a ≠ 1, N 1
, N
2
, N
3
, …, N
r
loga(N 1 .N 2 .N 3 ... Nr) = logaN 1 + logaN 2 + logaN 3 + …. + logaNr
(ii) (^) a
m log n
= log a
m – log a
n
Proof:
Given a > 0, a ≠ 1, m > 0, n > 0
Then suppose log a
m = x & log a
n = y
⇒ a
x = m & a
y = n
x
y
m a
n a
= = a
x – y
Change it into logarithmic form,
x – y = (^) a
m log n
⇒ logam – logan = (^) a
m log n
Q2. Solve log 2
10 – log 2
Sol. log 2 10 – log 2 5 = (^2)
log 5
(using P(2))
= log 2
(iii) logam
n = n logam
Proof:
Let’s say logam = α ⇔ a
α = m
then mn^ = (aα)n^ = anα
Change it into logarithmic form, then
nα = logam
n
nlog a
m = log a
mn
Logarithm 10.
Q5.
1023
2 n 1
log 1 = n
∑ is equal to:
(A) 8 (B) 9 (C) 10 (D) 12
Sol. C 1023
2 n 1
n 1 log = n
∑
log log log ... log 1 2 3 1023
+^ +^ +^ +
log.. ... 1 2 3 1023
= log 21024
= log 2
= 10log 2
Q6. Find the value of l 10 10 g 10 10
16log 12lo 7 log 15 2
og
+^ +
Sol. log 10 2 + 16log 10 16 – log 10 15 + 12log 10 25 – 12log 10 24 + 7log 10 81 – 7log 1080
= log 10
2 + 16log 10
24 – log 10
(3×5) + 12log 10
52 – 12log 10
(2^3 .3) + 7log 10
34 – 7log 10
(2^4.
= log 10 2 + 64log 10 2 – log 10 3 – log 3 5 + 24log 10 5 – 36log 102
- 12log 10 3 + 28log 10 3 – 28log 10 2 – 7log 105
= (1 + 64 – 36 – 28) log 10
2 + (–16 – 12 + 28) log 10
3 + (–16 – 24 – 7) log 10
= log 10 2 + log 105
= log 10
Base Changing theorem:
log b
a =
c
c
log a , log b
a > 0, b > 0, c > 0, a ≠ 1, b ≠ 1, c ≠ 1
Proof:
Suppose logba = x ⇔ a = b
x
logca = logcb
x
logca = xlogcb
c
c
log a x log b
= = log b
a
c b c
log a log a log b
(By taking logarithm)
11. Logarithm
Note:
a b
log b log a
Proof:
c
c c
c
log b 1
log a log^ a
log b
c c
c c
log b log b
log a log a
Hence proved.
(By using base Changing Theorem)
(Both are equal to each other)
Q7. Prove the following: logba. logcb. logdc = logda
Sol. log b
a. log c
b. log d
c =
log a log b log c log a
.. log b log c log d log d
= logda (Using Base-Changing Theorem)
Q8. If log 2
log 3
log 4
5...log n
(n + 1) = 10. Find ‘n’.
Sol. log 2 3. log 3 4. log 4 5...logn(n + 1) = 10
log 3 log 4 log 5 log(n 1)
... .... log 2 log 3 log 4 log n
log(n 1) 10 log 2
= (Using Base-Changing Theorem)
⇒ log 2 (n + 1) = 10
⇒ 210 = n + 1
⇒ n = 1024 –1 ⇒ n = 1023
Property of logarithm:
log c b log ab a =c
Proof:
a b a
log c log c log b a =a
( )
a (^) a
1 log c (^) log b = a
= ( ) a b
1 log b log a c =c
Hence b^ b
log c log a a =c
(Using Base-Changing theorem)
13. Logarithm
Q12. If plog 37 = 81 , then find value of
( )
2 log 3 7 p.
Sol. (^) ( ) ( )
log 3 7.log 3 7 log 3 7 log^3 7 log 3 7 p = p = 81
4 = 3 4log 3 7 = 3 log 37 = 74 = 2401
Q13. If a, b, c are real positive numbers such that 3 7 11
log 7 log 11 log 25 a = 27, b = 49, c = 11 ,
then find the value of
( log 3 7 ) 2 ( log 7 11 ) 2 ( log 11 25 )^2 a + b +c.
(A) 343 (B) 121 (C) 469 (D) 569
Sol. C
alog 3 7.log^3 7 + blog 7 11.log^7 11 +clog^11 25.log^11
= (^) ( ) ( ) ( )
log 3 7 log^3 7 log 7 11 log^7 11 log 11 25 log^11 a + b + c
= ( ) ( ) (^) ( )
log 3 7 log 7 11 log 11 25 27 + 49 + 11
3log 3 7 2log 7 11 ½log 1125 3 + 7 + 11
( ) 3 2 1/ log 3 7 log 7 11 log 11 25 3 + 7 + 11
= 7^3 + 11^2 + 5
Q14. If logax = b for permissible values of a and x then identify the statement(s) 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 23
First we will prove that log 2
3 is not rational
Suppose log 2 3 is rational, then log 2 3 = p/q ⇒ 3 = 2
p/q
⇒ 3 q^ = 2p
It is not possible for any value of p, q except p = 0 and q = 0
So, log 2 3 is irrational.
Hence (^2)
log 3 2
is irrational
So, we can say 2
log 3 will be irrational
Then ( ) =
log 23 2 3 is rational.
(B) a = 2, b = log 23
Then 2
log 3 2 = 3 is rational.
Logarithm 14.
(C) a = 3 , b = 2
( )
2 3 = 3 is rational.
(D) a = 2, b = 3
Then (2)^3 = 8 is rational.
Q15.
3 9 27
B
log 2 log 4 log 8
3 3
36 108
log 12 log 4 C log 3 log 3
Find B + C.
(A) 3 (B) 2 (C) 4 (D) 1
Sol. B
B = log 2 3 + 2log 4 9 – 3log 827
2 3
2 3 = log 3 2 + 2log 2 3 −3log 2 3
= log 2
3 + 2log 2
3 – 3log 2
3 3
36 108
log 12 log 4 C log 3 log 3
C = log 3 12 log 3 36 – log 3 4.log 3108
C = log 3 (4×3).log 3 (4×9) – log 3 4.log 3 (4×27)
C = (log 3
4 + log 3
3)(log 3
4 + log 3
- – log 3
4(log 3
4 + log 3
C = (log 3 4 + 1)(log 3 4 + 2) – log 3 4.(log 3 4 + 3)
Put log 3
4 = t,
C = (t + 1)(t + 2) – t(t + 3) = t
2
2
So, B + C = 0 + 2 = 2
Q16.
b b c c a a
A
1 log a log c 1 log a log b 1 log b log c
Where a > 0, a ≠ 1, b > 0, b ≠ 1, c > 0, c ≠ 1, abc ≠ 1, then A is :
(A) abc (B)
abc
(C) 1 (D) 0
Sol. C
b b b c c c a a a
log b log a log c log c log a log b log a log b log c
b c a
log abc log abc log abc
= log abc
a + log abc
c + log abc
a
= log abc abc = 1
Q17. Let a = log 3
5, b = log 17
25 and c = log 5
1000, d = log 7
2058 then which of
the following is/are true?
(A) a > b (B) a < b (C) c > d (D) c < d
Sol. AC
Logarithm 16.
Now,
3 3 (^ ) 3
(^25 ) 3 3 3
log 24 log^8 31 3log 2 log 24 log 25 log 5 2log 5
× +
By equation (ii), 2 + log 3
2 = β + 2βlog 3
⇒ (log 3 2)(1 – 2β) = β – 2
log 2 1 2
β −
− β
By equation (i), 1 + log 3
5 = α(1 + log 3
2 1 1 2
β − α (^) + −^ β
⇒ 1 + log 3
(^) − β + β − (^) − − β α (^) = α
−^ β^ −^ β
3
log 5 1 2
−αβ − α − + β
− β
Then
3 24 3
2 1 3 1 3log 2 1 2 log 25 2log 5 (^2 ) 2 1 2
(^) β −
- ^ − β = = (^) β − − αβ − α − β
( ) ( )
24
1 2 log
3 6 5 25 2 2 1 2 2 1
− β + β − β − = = β − − αβ − α β − − αβ − α
Q20. If log 7 12 = a and log 12 24 = b then find the value of log 54 168 in terms of a, b.
Sol. Given,
2 2 (^ ) 2
7 2 2 2
log 12 log^4 32 log 3 a log 12 log 7 log 7 log 7
× +
= = = = …(i)
and b = log 12 24 =
2 2 2
2 2 2
log 24 log^8 33 log 3
log 12 log 4 3 2 log 3
× +
× +
…(ii)
log 54 168 =
2 2 2 2
2 2 2
log 168 log^7 3 83 log 7 log 3
log 54 log 27 2 1 3log 3
× × + +
× +
… (iii)
From equation (ii), 3 + log 2 3 = 2b + b log 23
⇒ (1–b)log 2
3 = 2b – 3
2b 3 log 3 1 b
Replace this in equation (i),
2
2b 3 2 1 b log 7 a
2
2 2b 2b 3 log 7 a 1 b
2
log 7 a 1 b
17. Logarithm
Then from equation (iii),
54
2b 3 1 3 1 b a 1 b log 168 2b 3 1 3 1 b
54
3a 1 b a 2b 3 1 log 168 a 1 b 3 2b 3
[ ]
54
3a 3ab 2ab 3a 1 log 168 a 1 b 6b 9
( )
54
1 ab log 168 a 5b 8
Q21. If log 7 log 7 7 7 7
= 1 – a log 7
2 and log 15 log 15 15 15 15 15
= 1 – b log 15
Find a + b.
Sol. Let x log 7 log 7 7 7 7
7 7
x log log 7 7 7 2
7 ( 7 7 )
x log log 7 log 7 7 2
7 7
x log 1 log 7 7 2 2
^
7 7 7
x log 1 log 7 log 7 2 2 2
^ ^
7
x log 1 1 2 2 2
^ ^
7
x log 1. 2 2 2
^
7 7
x log log 2 4 8
^
x = 1 – log 7
x= 1 – 3log 72
Compare it with x = 1 – a log 7
2 then a = 3
Similarly, suppose y log 15 log 15 15 15 15 15
19. Logarithm
b ab b(^ c^ a)
log a k
= … (iv),
a ab c(^ a^ b)
log b k
= … (v)
Add equations (iv) & (v)
log (ab.ba) =
ab
k
[b + c – a + c + a –b] =
2abc
k
Consider the base as ‘e’,
we get,
2abc b a (^) k a b = e … (A)
Now multiply (ii) by c and (iii) by b, we get
bc c ( a b)
c log b k
= … (vi) and
bc a( b c)
b log c k
= … (vii)
Add (vi) and (vii), we get
( )
c b bc log b .c k
= [c +a – b + a + b – c] =
2abc
k
Change this into exponential form, we get 2abc c b (^) k b .c = e … (B)
Similarly we get,
2abc a c (^) k c .a = e … (C)
From equations (A), (B) and (C), we get
a
b .b
a = b
c .c
b = c
a .a
c
Logarithmic Equation
Q1. Find ‘x’ in following :
(i)^7
2 log x x + 7 − 2 = 0
Sol. Using a
log N a = N, we get
x
2
⇒ x
2
⇒ x(x + 2) – 1(x + 2) = 0
⇒ (x + 2) (x – 1) = 0
⇒ either x = 1 or x = –
Since x > 0, so we get x = 1
(ii)
log (^2) ( x^2 ) 2 − 3x − 4 = 0
Sol. Using property a
log N a =N
We get x
2 –3x – 4 = 0
⇒ x^2 – 4x + x – 4 = 0
⇒ x(x –4) + 1(x –4) = 0
⇒ (x –4)(x + 1) = 0
⇒ x = 4 and x = –
Logarithm 20.
(iii)
( )
x log 2 9 2 1 3 x
Sol. log 2
(9–2x) = 3 – x (change it into exponential form)
⇒ 9 – 2
x = 2
3–x
⇒ 9 – 2x^ = 2^3 .2–x^ = x
⇒ Put 2
x = t, we get
9 – t =
t
⇒ 9t – t^2 = 8
⇒ t
2
⇒ t
2
⇒ t(t –8) –1(t –8) = 0
⇒ t = 8, t = 1
⇒ 2 x^ = 2^3 and 2x^ = 1
⇒ x = 3 and x = 0, but x ≠ 3, hence x = 0
(iv) ( )
( )
log 10 x 1 x 1 100 x 1
Sol. By taking logarithm both sides with base 10, we get
( )
log 10 x 1 log 10 x 1 log 10 100 x 1
⇒ log 10
(x + 1).log 10
(x + 1) = log 10
100 + log 10
(x + 1)
Consider, log 10 (x + 1) = t, we get
t
2 = 2 + t
⇒ t^2 – t – 2 = 0
⇒ t
2
⇒ t(t –2) + 1(t –2) = 0
⇒ (t –2)(t + 1) = 0
⇒ t = 2, t = –
log 10
(x + 1) = 2 and log 10
(x +1) = –
⇒ x + 1 = 100 and (x + 1) =
⇒ x = 100 – 1 = 99 and
x 1 10
⇒ x = 99 and x =
⇒ x = 99 and x =
(v) log x–
4 = 1 + log 2
(x –1)
Sol. logx–1 2
2 = 1 + log 2 (x–1)
2 (^ )
2
1 log x 1 log x 1
