JEE Main Maths Logarithm Previous Year Questions with Solutions

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Download JEE Main Maths Logarithm Previous Year Questions With Solutions PDF

Question 1:

If (1/log₃π) + (1/log₄ π) > x, then x can be

(a) 3

(b) 2

(c) π

(d) None of the above

Solution:

Let y = (1/log₃π) + (1/log₄ π)

= (log 3/log π) + (log 4/log π)

= log_π 3 + log_π 4

= log_π 12 > x

So 12 > π^x

12 > π²

So, x = 2

Hence, option (b) is the answer.

Question 2:

The number of distinct solutions of the equation

Solution:

Given that

x ∈ [0, 2π]

⇒ log_1/2 |sin x| |cos x| = 2

⇒ |sin x| |cos x| = ¼

⇒ sin 2x = ±1/2

We have 8 solutions for x ∈ [0, 2π].

Question 3:

The value of

Solution:

(1/3)+(1/3²)+(1/3³)+….∞ = 1/3(1-1/3) = 1/2

log_2.5(1/2) ⇒ log_5/2 (1/2)

0.16 = 16/100 = 4/25 = (2/5)²

= 4

Question 4:

If

(a) 40

(b) 80

(c) 120

(d) 160

Solution:

Since,

Similarly,

and

Comparing, we get N = 120

Hence, option (c) is the answer.

Question 5:

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

⇒ ln 2 (ln 2x) = ln 3 (ln 3y)

⇒ ln 2 ln 2x = ln 3 (ln 3 + ln y) ..(i)

Also, 3^{ln x} = 2^{ln y}

⇒ ln x ln 3 = ln y ln 2

⇒ ln y = (ln x. ln 3)/ln 2 ..(ii)

Substitute (ii) in (i)

⇒ ln 2 (ln 2x) = ln 3 (ln 3 + (ln x. ln 3)/ln 2)

⇒ (ln 2)² (ln 2x) = ln 3 (ln 3 ln 2 + ln x. ln 3)

⇒ (ln 2)² (ln 2x) = (ln 3)² ( ln 2 + ln x)

⇒ (ln 2)² (ln 2x) = (ln 3)² ln 2x

⇒ [(ln 2)² – (ln 3)² ]ln 2x = 0

⇒ ln 2x = 0

⇒ 2x = 1

⇒ x = 1/2

Hence, option (c) is the answer.

Question 6:

If the sum of the first 20 terms of the series

is 460, then x is equal to:

(a) 7^1/2

(b) 7²

(c) e²

(d) 7^46/21

Solution:

460 = (2+3+4+…+21)log₇x

⇒ (20×(21+2)/2)log₇ x = 460

230 log₇x = 460

log₇x = 2

x = 7²

Hence, option (b) is the answer.

Question 7:

Logarithm of

(a) 3.6

(b) 5

(c) 4

(d) None of the above

Solution:

We use the property log_b aⁿ = n log_b a

32×4^1/5 = 2⁵× 2^2/5

= 2^27/5

2√2 = 2^3/2

= (27/5)(2/3) log₂ 2

= 18/5

= 3.6

Hence, option (a) is the answer.

Question 8:

If log₇ 2 = m, then log₄₉ 28 is equal to

(a) 2(1+2m)

(b) (1+2m)/2

(c) 2/(1+2m)

(d) 1 + m

Solution:

Given log₇ 2 = m

= (1/2) log₇ 28

= (1/2) [log₇ (4×7)]

= (1/2) [log₇ 4+ log₇ 7)]

= (1/2) [log₇ 2² + 1)]

= (1/2) [ 2 log₇ 2 + 1)]

= (1/2) [2m + 1)]

Hence, option (b) is the answer.

Question 9:

The value of log₃ 4 log₄ 5 log₅ 6 log₆ 7 log₇ 8 log₈ 9 is

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

We know log_b a = log a/log b

So log₃ 4 log₄ 5 log₅ 6 log₆ 7 log₇ 8 log₈ 9 = (log 4/log 3) × (log 5/log 4) × (log 6/log 5) ×

(log 7/log 6) × (log 8/log 7) × (log 9/log 8)

= log 9/log 3

= log 3²/log 3

= 2 log 3 /log 3

= 2

Hence, option (b) is the answer.

Question 10:

If a² + 4b² = 12ab, then log(a + 2b) is

(a) ½ (4 log 2 + log a + log b)

(b) 0

(c) (log 2 + log a + log b)

(d) (4 log 2 + log a – log b)

Solution:

Given, a² + 4b² = 12ab

Adding 4ab on both sides, we get

a² + 4b² + 4ab = 16ab

⇒ (a + 2b)² = 16ab

Taking log on both sides,

log (a + 2b)² = log (16ab)

2 log (a + 2b) = log 16 + log a + log b

2 log (a + 2b) = log 2⁴ + log a + log b

2 log (a + 2b) = 4 log 2 + log a + log b

log (a + 2b) = ½ (4 log 2 + log a + log b)

Hence, option (a) is the answer.

Question 11:

The value of

(a) 81

(b) 1/81

(c) 20

(d) 1/20

Solution:

We use the following properties in this problem.

log_b aⁿ = n log_b a

Here 0.1 + 0.01 + 0.001 + … = 0.1/(1 – 0.1) (use sum of infinite GP formula)

= 1/9

= 2 log₂₀(1/9)

= 2 log₂₀(9^-1)

= -2 log₂₀ 9

= 81

Hence, option (a) is the answer.

Question 12:

If a, b, c are distinct positive numbers, each different from 1, such that [log_b a log_c a – log_a a] + [log_a b log_c b – log_b b] + [log_a c log_b c – log_c c] = 0, then abc =

(a) 1

(b) 2

(c) 3

(d) 4

Solution:

We use the property log_n m = log m/log n

Given [log_b a log_c a – log_a a] + [log_a b log_c b – log_b b] + [log_a c log_b c – log_c c] = 0

⇒ [(log a/log b)(log a/log c) – 1] + [(log b/log a)(log b/log c) – 1] + [(log c/log a)(log c/log b) – 1] = 0

⇒ [(log a)²/log b log c] + [(log b)²/log a log c] + [(log c)²/log a log b] = 3

Multiply both sides by log a log b log c

⇒ (log a)³ + (log b)³ + (log c)³ = 3 log a log b log c

⇒ log a + log b + log c = 0 (Use x³ + y³ + z³ = 3xyz, then x + y + z = 0)

⇒ log (abc) = 0

⇒ abc = 1

Hence, option (a) is the answer.

Question 13:

If a, b, and c are consecutive positive integers and log(1 + ac)= 2k, then the value of k is

(a) log a

(b) log b

(c) 2

(d) 1

Solution:

Let a = n-1, b = n, c = n+1, where n > 1.

Given log (1 + ac) = 2k

⇒ log (1 + (n-1)(n+1)) = 2k

⇒ log (1 + n² – 1) = 2k

⇒ log n² = 2k

⇒ 2log n = 2 k

⇒ log n = k

⇒ log b = k

Hence, option (2) is the answer.

Question 14:

If log₃ x + log₃ y = 2 + log₃ 2 and log₃(x + y) = 2, then

(a) x = 1, y = 8

(b) x = 8, y = 1

(c) x = 3, y = 6

(d) x = 9, y = 3

Solution:

Given log₃ x + log₃ y = 2 + log₃ 2

⇒ log₃ xy = log₃3² + log₃ 2

⇒ log₃ xy = log₃9 + log₃ 2

⇒ log₃ xy = log₃18 (since log a + log b = log ab)

⇒xy = 18…(i)

log₃(x + y) = 2

⇒ x + y = 3²

⇒ x + y = 9…(ii)

Solving (i) and (ii), we get,

x = 3, y = 6

Hence, option (c) is the answer.

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