Logarithms CAT Previous Year Questions with Answer PDF

2024

Q1: If x is a positive number such that 4 log₁₀x + 4 log₁₀₀x + 8 log₁₀₀₀x = 13, then the greatest integer not exceeding x, is 

Ans: 31

Sol: 

Q2: If a, b and c are postive real numbers such that a > 10 ≥ b ≥ c and then the greatest possible integer value of a is 

Ans: 14
Sol: 

The value of a will be maximum, when both b and c will be maximum which is equal to 10 but, in that case a² will be equal to 200 which is not possible for the integer value of a.
So, the greatest possible value a² = 196.
So, the greatest possible integer value of a = 14
Hence, 14 is the required answer.

Q3: If 3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8 and 8^f = 9, then the value of the product abcdef is 

Ans: 2
Sol: We have
3^a = 4 => a = log₃ 4
4^b = 5 => b = log₄ 5
5^c = 6 => c = log₅ 6
6^d = 7 => d = log₆ 7
7^e = 8 => e = log₇ 8
8^f = 9 => f = log₈ 9
Now, a * b * c * d * e * f = log₃ 4 * log₄ 5 * … * log₈ 9 = log₃ 9 / log₃ 3 = 2
Hence, 2 is the required answer.

2023

Q1: If x and y are positive real numbers such that log_x (x² + 12) = 4 and 3 log_y x = 1 , then x + y equals  [2023]
(a) 20
(b) 11
(c) 68
(d) 10

Ans: d

Sol: log_x (x² + 12) = 4
This means, x² + 12 = x⁴
Let k = x²
k + 12 = k²
k² – k – 12 = 0
k² – 4k + 3k – 12 = 0k(k – 4) + 3(k – 4) = 0k = 4 or k = -3
But since k = x², k is always non-negative.∴ k = 4
∴ x² = 4Since x is the base of the log function, it can should always be positive.
∴ x = 2
3 log_yx = 1
log_yx³ = 1
x³ = y¹
y = x³ = 2³ = 8
∴ x + y = 2 + 8 = 10

Q2: For some positive real number , then the value of log₃ (3x²) is  [2023]

Ans: 7

Sol: 

Q3: For a real number x, if are in an arithmetic progression, then the common difference is  [2023
(a) log₄7
(b) log₄(3/2)
(c) log₄(7/2)
(d) log₄(23/2)

Ans: c

Sol: 

But, since the argument of the log can't be negative, x = 4

∴ The common ratio of the G.P is 

∴ The common difference of the AP = (7/2)

Q4: The sum of all possible values of x satisfying the equation ₂4x² - ₂2x² + x + 16 + 2^{2x + 30} = 0, is 
(a) 3/2
(b) 5/2
(c) 1/2
(d) 3

Ans: c
Sol: 

2022

Q1: The number of distinct integer values of n satisfying is  [2022]

Ans: 47

Sol: 

For any value of n < 16, the numerator is positive. For any value of n > 16, it is negative.For any value of n < 64, the denominator is positive. For any value of n > 64, it is negative.For a fraction to be negative, the numerator and the denominator must be of opposite signs.In this case, n should be between 16 and 64.

The number of values that n can take = 63 - 16 = 47.

Q2: If  for all non-zero real values of a and b, then the value of x + y is 

Ans: 14
Sol:
(7/5)^(3x-y)/2 = 875/2401 = 125/343 = (7/5)^-3 
=> y – 3x = 6
(4a/b)^6x-y = (b/2a)^6x-y
=> But 4a/b ≠ b/2a
=> The power is 0.
=> y = 6x
=> x = 2, y = 12
x + y = 14

2021

Q1: For a real number a, if  then a must lie in the range  [2021]
(a) 2 < a < 3
(b) 3 < a < 4 
(c) 4 < a < 5
(d) a > 5

Ans: c

Sol: 

Q2: If  = 0 then 4x equals  [2021]

Ans: 5

Sol: We have :

4x = 5

Q3: If ,  then 100x equals  [2021]

Ans: 99

Sol: We can re-write the equation as:

Squaring both sides:

Hence,

2020

Q1: If log₄ 5 = log₄ y log₆ √5 , then y equals  [2020]

Ans: 36

Sol: 

⇒ y = 36

Q2: If y is a negative number such that 2^{y2log₃ 5} = 5^{log₂ 3} , then y equals  [2020]
(a)  log2(1 / 3)
(b)  - log2(1 / 3)
(c)  log2(1 / 5)
(d) - log2(1 / 5)

Ans: a

Sol: 2^{y2log₃ 5} = 5^{log₂ 3}

( is negative )

Q3: The value of , for 1 < a ≤ b cannot be equal to  [2020] 
(a)  -0.5
(b)  1
(c)  0
(d) -1

Ans: b

Sol: 
= log_a a - log_ab +log_bb - log_ba
=
since ( log_ab + log_ab ) ≥ 2
∴ 1 can't be the answer.

Q4:  equals  [2020]

Ans: 24

Sol: 
= 24

Q5: In the final examination, Bishnu scored 52% and Asha scored 64%. The marks obtained by Bishnu is 23 less, and that by Asha is 34 more than the marks obtained by Ramesh. The marks obtained by Geeta, who scored 84%, is  [2020]
(a)  469
(b)  399
(c)  357
(d) 417

Ans: b

Sol: Let the total marks be T and scores of Bishnu, Asha and Ramesh be a, b and c respectively. Given, a = 52% of T = c - 23 and b = 64% of T = c + 34Hence, (64 - 52)% of T = (c + 34) - (c - 23) = 57i.e. 12% of T = 57
Hence, score of Geeta = 84% of T = 7  x 57 = 399

2019

Q1: Let x and y be positive real numbers such that log₅(x + y) + log₅(x - y) = 3, and log₂y - log₂x = 1 - log₂3. Then xy equals  [2019]
(a) 25
(b) 150
(c) 250
(d) 100

Ans: b

Sol: Given that, log₅(x + y) + log₅(x - y) = 3We know log A + log B = log (A x B)
log₅(x + y) + log₅(x - y) = log₅(x² - y²)
log₅(x² - y²) = 3
x² - y² = 5³
x² - y² = 125
Similarly, log₂y - log₂x = 1 - log₂3
log₂ (y/x) = log₂2 - log₂3
log₂(y/x) = log₂(2/3)3y = 2x
(3/2)y² - y² = 125
(9/4)y² - y² = 125
5/4 y² = 125
y² = 100y = 10x = 15So, xy = 15 x 10 = 150

Q2: If x is a real number ,then  is a real number if and only if  [2019]
(a) -3 ≤ x ≤ 3
(b) 1 ≤ x ≤ 2
(c) 1 ≤ x ≤ 3
(d) -1 ≤ x ≤ 3

Ans: c

Sol: It is given that, is a real number
Therefore, ≥ 0
≥ 1
4x - x² ≥ 3
x² - 4x + 3 ≤ 0(x−1) (x-3) ≤ 0x ∈ [1,3]

Q3: The real root of the equation 2^6x + 2^3x+2 - 21 = 0 is  [2019]
(a) 
(b) log₂9
(c) 
(d) log₂27

Ans: a

Sol: Given equation - 2^6x + 2^3x+2 - 21 = 0The idea is to convert the above equation to a quadratic one.
So, 2^6x = (2^3x)²
Let y = 2^3x
Now, 2^6x + 2^3x+2 - 21 = 0 can be rewritten as + 2^(3x)2+ 2².2^3x - 21 = 0
y² + 4y - 21 = 0Solving the above quadratic equation,(y + 7) (y - 3) = 0So, y = -7 or +3
y = 2^3x ⇒ y should always be positiveTherefore, y = -7 is not a valid solution. So, only y = +3 exists.
2^3x = 3Taking log on both sides,
log₂ 2^3x = log₂ 3
3x = log₂ 3
x = 1/3 log₂ 3, which is the real solution to the given equation

2018

Q1: If x is a positive quantity such that 2^x = 3^log52 , then x is equal to  [2018]
(a) log₅9
(b) 1 + log₅(3/5)
(c) 1 + log₃ (5/3)
(d) log₅8

Ans: b

Sol: Given, 2^x = 3^log52Taking log on both sides, we get
log (2^x) = log (3^log52)
x × log(2) = log₅(2) × log(3)
Let us consider the base as 3,
Going through the options, Simplifying (B)
1 + log₅ (3/5)
1 + log₅(3) - log₅(5)
1 + log₅(3) – 1
log₅(3)

Q2: If log₂(5 + log₃a) = 3 and log₅(4a + 12 + log₂b) = 3, then a + b is equal to  [2018]
(a) 32
(b) 59
(c) 67
(d) 40

Ans: b

Sol: We know that if logqp = r then p = q^r
Given log₂(5 + log₃a) = 3
5 + log₃a = 23
log₃a = 3
a = 3³ = 27
Similarly, 4a + 12 + log₂b = 5³
4a + log₂b = 113
log₂b = 5
b = 2⁵ = 32
a + b = 27 + 32 = 59

Q3: If log₁₂81 = p, then  is equal to:  [2018]
(a) log₂8
(b) log₆8
(c) log₄16
(d) log₆16

Ans: b

Sol: Given, log₁₂81 = p
log₁₂(3)⁴ = p
4 (log₁₂(3)) = p
log₁₂(3) = p/4 ...(1)
We need to find the value of 3 ×
This can be rewritten as 3 × Replacing (2) with (1), We get




log₃₆64
log₆8
Hence, the answer is log₆8

Q4: If p³ = q⁴ = r⁵ = s⁶, then the value of logs (pqr) is equal to  [2018]
(a) 24/5
(b) 1
(c) 47/10
(d) 16/5

Ans: c

Sol: Given that p³ = q⁴ = r⁵ = s⁶
We have to find the value of log_s (pqr)Since more variables are given, to avoid confusion assume a new variable to simplify it.
Let us assume this p³ = q⁴ = r⁵ = s⁶ is equal to k^xso that we can get every values in k
We can rewrite this logs (pqr) as Now let us take the LCM of 3 , 4 , 5 and 6 which is equal to 60
Hence p³ = q⁴ = r⁵ = s⁶ = k⁶⁰
Or p = k²⁰ q = k¹⁵ r = k¹² s = k¹⁰

The question is "If p³ = q⁴ = r⁵ = s⁶, then the value of logs (pqr) is equal to" Hence, the answer is 47/10

Q5:    [2018]
(a) 0
(b) 1/2
(c) -4
(d) 10

Ans: b

Sol: We know that
⇒log₁₀₀ 2 - log₁₀₀ 4 + log₁₀₀ 5 - log₁₀₀ 10 + log₁₀₀ 20 - log₁₀₀ 25 + log₁₀₀ 50
We also know that (- log_a b) = log_a (1/b)

2017

Q1: Suppose, log₃x = log₁₂y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log₆G is equal to:  [2017]
(a) √a
(b) 2a
(c) a/2
(d) a

Ans: d

Sol: log₃ x = log₁₂ y = a, where x, y are positive numbers.
log₃ x = a; x = 3^a
log₁₂ y = a; y = 12^a
If G is the geometric mean of x and y, log₆G is equal to has to be foundWe know that geometric mean is √xy
√xy = √(36^a)
= √(6^2a) = (6^2a)^1/2
= 6^a
log₆ G = a

Q2: If 9^{2x – 1} – 81^x-1 = 1944, then x is   [2017]
(a) 3
(b) 9/4
(c) 4/9
(d) 1/3

Ans: b

Sol: We have to find the value of x
⇒ 9^{2x – 1} – 81^x-1 = 1944
⇒ 9^{2x – 1} – (9²)^x-1 = 1944
⇒ 9^{2x – 1} – 9^{2x - 2} = 1944
⇒ (9^{2x - 2} × 9) - 9^{2x - 2} = 1944

We can write 243 = 3⁵ = 9^(5/2)Comparing the powers,⇒ 2x – 2 = 5/2⇒ 2x = 9/2⇒ x = 9/4

Q3: If x is a real number such that log₃5 = log₅(2 + x), then which of the following is true?   [2017]
(a) 0 < x < 3
(b) 23 < x < 30
(c) x > 30
(d) 3 < x < 23

Ans: d
Sol: log₃ 5 = log₅ (2 + x)
Since log3 3 = 1 and log₃ 9 = 2 , we can say that 2 > log₃ 5 > 1We need to find the range of x
If log₅ (2 + x) > 1⇒ 2 + x > 5⇒ x > 3
If log₅ (2 + x) < 2
⇒ 2 + x < 5²⇒ 2 + x < 25⇒ x < 23
Therefore 3 < x < 23

Q4: If log(2^a × 3^b × 5^c) is the arithmetic mean of log(2² × 3³ × 5), log(2⁶ × 3 × 5⁷), and log(2 × 3² × 5⁴), then a equals   [TITA 2017]

Ans: 3

Sol: 3log(2^a × 3^b × 5^c) = log(2² × 3³ × 5) + log(2⁶ × 3 × 5⁷) + log(2 × 3² × 5⁴)We have to find only the value of a
2^3a × 3^3b × 5^3c = (2² × 2⁶ × 2¹) × (3³ × 3 × 3²) × (5 × 5⁷ × 5⁴)
2^3a = 2⁹3a = 9a = 3

Q5: The value of log_0.008√5 + log_√381 – 7 is equal to:  [2017]
(a) 1/3
(b) 2/3
(c) 5/6
(d) 7/6

Ans: c

Sol: Here we have to find the value of log_0.008 √5 + log_√3 81 – 7.
⟹ log_0.008 √5 + log_√3 81 – 7
log_0.008 √5 can be written in the terms of five and log_√3 81 can be written in the terms of 3.
where √5 = 5^1/2 ⟹ 0.008 = 2³/10³ = 5^-3
⟹ log_0.008 √5 + log_√3 81 – 7

Hence the value of log_0.008√5 + log_√381 – 7 = 5/6

Q6: If 9^{x - (1/2)} – 2^{2x – 2} = 4^x – 3^{2x – 3}, then x is  [2017]
(a) 3/2
(b) 2/5
(c) 3/4
(d) 4/9

Ans: a

Sol: Given that

⇒ 3^2x × 8 = 2^2x × 27
⇒ 3^{2x - 3} = 2^{2x - 3}⇒ 2x - 3 = 0⇒ x = 3/2