JEE Advanced Maths Logarithm Previous Year Questions with Solutions

Logarithm JEE Advanced previous year questions with solutions are given on this page. We define logarithm as the inverse function of exponentiation. These solutions are prepared by our subject experts with the aim to help you perform exceedingly well in the JEE Advanced exam. Students are recommended to revise and learn these solutions so that they will be familiar with the type of questions asked from this topic. This will also help them to be stress-free during the exam.

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JEE Advanced Maths Logarithm Previous Year Questions with Solutions

Question 1: Let (x₀, y₀) be the solution of the following equations.

(2x)^{ln 2} = (3y)^{ln 3}

3^{ln x} = 2^{ln y}. Then x₀ is

(a) 1/6

(b) 1/3

(c) 1/2

(d) 6

Solution:

Given that (2x)^{ln 2} = (3y)^{ln 3} ..(i)

Taking log on both sides

log 2 log (2x) = log 3 log (3y)

log 2 (log 2 + log x) = log 3 (log 3 + log y) ..(ii)

Also 3^{ln x} = 2^{ln y} ..(iii)

Taking log on both sides

log x log 3 = log y log 2

log y = log x log 3/log 2 …(iv)

Substitute (iv) in (ii)

log 2 (log 2 + log x) = log 3 (log 3 + log x log 3/log 2 )

(log 2)² + log 2 log x = (log 3)² + log x (log 3)²/log 2

log x = [(log 3)² – (log 2)²]/[ ((log 2)² – (log 3)² )/log 2]

log x = – log 2

log x = log 2^-1

=> x = 2^-1

= 1/2

Hence option c is the answer.

Question 2: If 3^x = 4^x-1, then x =

Solution:

Given that 3^x = 4^x-1

Take log on both sides

log 3^x = log 4^x-1

x log 3 = (x-1) log 4

= x log 4 – log 4

log 4 = x(log 4 – log 3)

So x = log 4/(log 4 – log 3)

= 1/(1 – log 3/log 4)

= 1/(1 – log₄ 3)

Hence the value of x is 1/(1 – log₄ 3).

Question 3: The number of positive integers satisfying the equation x + log₁₀(2^x + 1) = x log₁₀5 + log₁₀ 6 is

(a) 0

(b) 1

(c) 2

(d) infinite

Solution:

x + log₁₀(2^x + 1) = x log₁₀5 + log₁₀ 6

x [1 – log₁₀5] + log₁₀(2^x + 1) = log₁₀ 6

x [ log₁₀ 10 – log₁₀ 5] + log₁₀(2^x + 1) = log₁₀ 6

x log₁₀2 + log₁₀(2^x + 1) = log₁₀ 6

log₁₀ 2^x (2^x + 1) = log₁₀ 6

(2^x)² + 2^x – 6 = 0

2^x = 2 or 2^x = -3 (rejected)

=> x = 1

So number of positive integers = 1

Hence option b is the answer.

Question 4: If log₇ 2 = k, then log₄₉ 28 is equal to

(a) (1+2k)/4

(b) (1+2k)/2

(c) (1+2k)/3

(d) none of the above

Solution:

Given log₇ 2 = k

log₄₉ 28 =

= ½ log₇ 28

= ½ log₇ (4×7)

= ½ log₇ 4 + ½ log₇ 7

= ½ log₇ 2² + ½

= log₇ 2 + ½

= k + ½

= (2k+1)/2

Hence option b is the answer.

Question 5: If log_a x = b for permissible values of a and x then which is/are correct.

(a) If a and b are two irrational numbers, then x can be rational

(b) If a is rational and b is irrational, then x can be rational

(c) If a is irrational and b is rational, then x can be rational

(d) If a and b are rational, then x can be rational

Solution:

Given that log_a x = b

=> x = e^b/a

Case 1: Let, b be any irrational number and a be equal to a = e^b that means a is irrational because e is irrational. So x = e^b/a becomes rational.

Hence option a is correct.

Case 2: Let a be rational and b is irrational.

If we take any rational number as a and b = ln a, then x = e^b/a = e^{ln a}/a = a/a = 1.

So x is rational.

Hence option b is correct.

Case 3: Let a be irrational and b is rational,

If we consider any rational number as b and a = e^b

Then x = e^b/e^b = 1 is rational.

Hence option c is correct.

Case 4: let a and b be rational.

If we take a = any rational number and b = 0, x can be rational.

Hence option d is correct.

So option a, b, c and d are correct.

Question 6: If x = 9 is a solution of ln(x² + 15a²) – ln (a-2) = ln (8ax/(a-2)) then

(a) a = 3/5

(b) a = 3

(c) x = 15

(d) x = 2

Solution:

Given ln(x² + 15a²) – ln (a-2) = ln (8ax/(a-2))

ln [(x² + 15a²)/(a-2)] = ln (8ax/(a-2))

=> (x² + 15a²)/(a-2) = (8ax/(a-2))

=> x² + 15a² = 8ax

=> x² + 15a² – 8ax = 0 ..(i)

Given x = 9 is a root.

=> 81 + 15a² – 72a = 0

=> 5a² – 24a + 27 = 0

=> (5a-9) (a-3) = 0

=> a = 9/5 or a = 3

Put value of a in (i)

When a = 9/5, we get x = 9 or x = 27/5

When a = 3, we get x = 9 or x = 15

Hence option b and c are correct.

Question 7: If log₁₀ 5 = a and log₁₀ 3 = b then

(a) log₃₀ 8 = 3(1-a)/(b+1)

(b) log₄₀ 15 = (a+b)/(3-2a)

(c) log₂₄₃ 32 = (1-a)/b

(d) none of these

Solution:

Given log₁₀ 5 = a and log₁₀ 3 = b

We check all the given options.

log₃₀ 8 = log 2³/log (3×10)

= 3 log 2/( log 3 + log 10)

= 3 log (10/5)/(1 + log 3)

= 3 (1 – log 5)/(1 + log 3)

= 3(1 – a)/(1+b)

Hence option a is correct.

log₄₀ 15 = log 15/log 40

= log(3×5)/log(10×4)

= (log 3 + log 5)/(1 + log 2²)

= (log 3 + log 5)/(1 +2 log 2)

= (log 3 + log 5)/(1 +2 log (10/5))

= (log 3 + log 5)/(1 +2 (1 – log 5))

= (log 3 + log 5)/(1 +2 – 2 log 5)

= (b+a)/(3 – 2a)

Hence option b is correct.

log₂₄₃ 32 = log 32/log 243

= log 2⁵/log 3⁵

= 5 log 2/5 log 3

= log 2/log 3

= (1 – log 5)/log 3 (since log 2 = log (10/5) = log 10 – log 5 = 1 – log 5)

= (1-a)/b

So option c is correct.

Hence option a, b and c are correct.

Question 8: The value of

(a) 8

(b) 1

(c) 2

(d) none of these

Solution:

Hence option a is the answer.

Question 9: If log₁₀ 2 = a and log₁₀ 3 = b then log 60 can be expressed in terms of a and b as

(a) a+b+1

(b) a-b+1

(c) a-b-1

(d) a+b-1

Solution:

Given log₁₀ 2 = a and log₁₀ 3 = b

log 60 = log (10×2×3)

= log 10 + log 2 + log 3

= 1 + a + b

Hence option a is the answer.

Question 10: If A = log₂ log₂ log₄ 256 + log_√2 2, then A is equal to

(a) 2

(b) 3

(c) 5

(d) 7

Solution:

We use the property log_a (m)ⁿ = n log_a m

log₂ log₂ log₄ 256 + log_√2 2 = log₂ log₂ log₄ (4)⁴ + log_√2 √2²

= log₂ log₂ 4 + 2(2)

= log₂ log₂ 2² + 4

= log₂ 2 + 4

= 1+4 (log_a a = 1)

= 5

Hence option c is the answer.

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