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Logarithm: JEE Main Previous Year Questions (2021-2026)
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Page 1
JEE Main Previous Year Questions (2021-2026):
Logarithm
(January 2026)
Q1: The sum of all the real solutions of the equation log
(x+3)
(6x
2
+ 28x + 30) = 5 - 2log
(6x +
10)
(x
2
+ 6x + 9) is equal to :
(a) 1
(b) 4
(c) 0
(d) 2
Ans: (c)
Sol:
Checking x in equation (1)
Page 2
JEE Main Previous Year Questions (2021-2026):
Logarithm
(January 2026)
Q1: The sum of all the real solutions of the equation log
(x+3)
(6x
2
+ 28x + 30) = 5 - 2log
(6x +
10)
(x
2
+ 6x + 9) is equal to :
(a) 1
(b) 4
(c) 0
(d) 2
Ans: (c)
Sol:
Checking x in equation (1)
Checking x in equation (1)
So, solutions are x = 1 and x = -1
Sum of solution = 1 + (-1) = 0
Page 3
JEE Main Previous Year Questions (2021-2026):
Logarithm
(January 2026)
Q1: The sum of all the real solutions of the equation log
(x+3)
(6x
2
+ 28x + 30) = 5 - 2log
(6x +
10)
(x
2
+ 6x + 9) is equal to :
(a) 1
(b) 4
(c) 0
(d) 2
Ans: (c)
Sol:
Checking x in equation (1)
Checking x in equation (1)
So, solutions are x = 1 and x = -1
Sum of solution = 1 + (-1) = 0
Q1: The product of all solutions of the equation ?? ?? ( ???? ?? ?? ?? )
?? + ?? = ?? ?? , ?? > ?? , is :
(A) ?? 2
(B) ??
(C) ?? 6
5
(D) e
8 / 5
Ans: (D)
Solution:
We begin with the equation:
?? 5 ( l og
?? ? ?? )
2
+ 3
= ?? 8
, ?? > 0
Equating the exponents, we have:
5 ( log
?? ? ?? )
2
+ 3 = log
?? ? ?? 8
= 8 log
?? ? ??
Let ?? = log
?? ? ?? . Substituting, the equation becomes:
5 ?? 2
+ 3 = 8 ??
Rewriting this as a quadratic equation:
5 ?? 2
- 8 ?? + 3 = 0
Factoring the quadratic:
5 ?? 2
- 5 ?? - 3 ?? + 3 = 0
? ( 5 ?? - 3 ) ( ?? - 1 ) = 0
Thus, the solutions for ?? are:
?? = 1 implies log
?? ? ?? = 1, giving ?? = ?? .
?? =
3
5
implies log
?? ? ?? =
3
5
, giving ?? = ?? 3
5
.
The product of all solutions is:
?? 1
× ?? 3
5
= ?? 1 +
3
5
= ?? 8
5
Page 4
JEE Main Previous Year Questions (2021-2026):
Logarithm
(January 2026)
Q1: The sum of all the real solutions of the equation log
(x+3)
(6x
2
+ 28x + 30) = 5 - 2log
(6x +
10)
(x
2
+ 6x + 9) is equal to :
(a) 1
(b) 4
(c) 0
(d) 2
Ans: (c)
Sol:
Checking x in equation (1)
Checking x in equation (1)
So, solutions are x = 1 and x = -1
Sum of solution = 1 + (-1) = 0
Q1: The product of all solutions of the equation ?? ?? ( ???? ?? ?? ?? )
?? + ?? = ?? ?? , ?? > ?? , is :
(A) ?? 2
(B) ??
(C) ?? 6
5
(D) e
8 / 5
Ans: (D)
Solution:
We begin with the equation:
?? 5 ( l og
?? ? ?? )
2
+ 3
= ?? 8
, ?? > 0
Equating the exponents, we have:
5 ( log
?? ? ?? )
2
+ 3 = log
?? ? ?? 8
= 8 log
?? ? ??
Let ?? = log
?? ? ?? . Substituting, the equation becomes:
5 ?? 2
+ 3 = 8 ??
Rewriting this as a quadratic equation:
5 ?? 2
- 8 ?? + 3 = 0
Factoring the quadratic:
5 ?? 2
- 5 ?? - 3 ?? + 3 = 0
? ( 5 ?? - 3 ) ( ?? - 1 ) = 0
Thus, the solutions for ?? are:
?? = 1 implies log
?? ? ?? = 1, giving ?? = ?? .
?? =
3
5
implies log
?? ? ?? =
3
5
, giving ?? = ?? 3
5
.
The product of all solutions is:
?? 1
× ?? 3
5
= ?? 1 +
3
5
= ?? 8
5
2023
Q1: Let ?? , ?? , ?? be three distinct positive real numbers such that ( ?? ?? )
?? ?? ?? ?? ? ?? = ( ???? )
?? ?? ?? ?? ? ?? and ?? ?? ?? ?? ?? ? ?? =
?? ?? ?? ?? ?? ? ?? .
Then, ?? ?? + ?? ???? is equal to [JEE Main 2023 (Online) 10th April Morning Shift]
Ans: 8
Given, ( 2 ?? )
ln ? ?? = ( ???? )
ln ? ?? , where 2 ?? > 0 , ???? > 0
? ln ? ?? ( ln ? 2 + ln ? ?? ) = ln ? ?? ( ln ? ?? + ln ? ?? ) .
and ( ?? )
ln ? 2
= ( ?? )
ln ? ??
? ln ? 2 · ln ? ?? = ln ? ?? · ln ? ?? .
Now, let ln ? ?? = ?? , ln ? ?? = ??
ln ? 2 = ?? , ln ? ?? = ??
Now, from Eqs. (i) and (ii), we get
?? · ?? = ???? and ?? ( ?? + ?? ) = ?? ( ?? + ?? )
? ?? =
????
??
? ? ?? (
????
?? + ?? ) = ?? ( ?? + ?? )
? ? ?? 2
?? + ?? 2
?? = ?? 2
( ?? + ?? )
? ? ( ?? 2
- ?? 2
) ( ?? + ?? ) = 0
? ? ?? 2
= ?? 2
?? = ± ??
So, from the equations, there are two cases:
Case ?? :
?? = ??
In this case, since ?? and ?? are natural logarithms of positive numbers ?? and ?? respectively, this implies
that ?? = ?? . However, this cannot be true as a, ?? , and ?? are given to be distinct positive real numbers.
Page 5
JEE Main Previous Year Questions (2021-2026):
Logarithm
(January 2026)
Q1: The sum of all the real solutions of the equation log
(x+3)
(6x
2
+ 28x + 30) = 5 - 2log
(6x +
10)
(x
2
+ 6x + 9) is equal to :
(a) 1
(b) 4
(c) 0
(d) 2
Ans: (c)
Sol:
Checking x in equation (1)
Checking x in equation (1)
So, solutions are x = 1 and x = -1
Sum of solution = 1 + (-1) = 0
Q1: The product of all solutions of the equation ?? ?? ( ???? ?? ?? ?? )
?? + ?? = ?? ?? , ?? > ?? , is :
(A) ?? 2
(B) ??
(C) ?? 6
5
(D) e
8 / 5
Ans: (D)
Solution:
We begin with the equation:
?? 5 ( l og
?? ? ?? )
2
+ 3
= ?? 8
, ?? > 0
Equating the exponents, we have:
5 ( log
?? ? ?? )
2
+ 3 = log
?? ? ?? 8
= 8 log
?? ? ??
Let ?? = log
?? ? ?? . Substituting, the equation becomes:
5 ?? 2
+ 3 = 8 ??
Rewriting this as a quadratic equation:
5 ?? 2
- 8 ?? + 3 = 0
Factoring the quadratic:
5 ?? 2
- 5 ?? - 3 ?? + 3 = 0
? ( 5 ?? - 3 ) ( ?? - 1 ) = 0
Thus, the solutions for ?? are:
?? = 1 implies log
?? ? ?? = 1, giving ?? = ?? .
?? =
3
5
implies log
?? ? ?? =
3
5
, giving ?? = ?? 3
5
.
The product of all solutions is:
?? 1
× ?? 3
5
= ?? 1 +
3
5
= ?? 8
5
2023
Q1: Let ?? , ?? , ?? be three distinct positive real numbers such that ( ?? ?? )
?? ?? ?? ?? ? ?? = ( ???? )
?? ?? ?? ?? ? ?? and ?? ?? ?? ?? ?? ? ?? =
?? ?? ?? ?? ?? ? ?? .
Then, ?? ?? + ?? ???? is equal to [JEE Main 2023 (Online) 10th April Morning Shift]
Ans: 8
Given, ( 2 ?? )
ln ? ?? = ( ???? )
ln ? ?? , where 2 ?? > 0 , ???? > 0
? ln ? ?? ( ln ? 2 + ln ? ?? ) = ln ? ?? ( ln ? ?? + ln ? ?? ) .
and ( ?? )
ln ? 2
= ( ?? )
ln ? ??
? ln ? 2 · ln ? ?? = ln ? ?? · ln ? ?? .
Now, let ln ? ?? = ?? , ln ? ?? = ??
ln ? 2 = ?? , ln ? ?? = ??
Now, from Eqs. (i) and (ii), we get
?? · ?? = ???? and ?? ( ?? + ?? ) = ?? ( ?? + ?? )
? ?? =
????
??
? ? ?? (
????
?? + ?? ) = ?? ( ?? + ?? )
? ? ?? 2
?? + ?? 2
?? = ?? 2
( ?? + ?? )
? ? ( ?? 2
- ?? 2
) ( ?? + ?? ) = 0
? ? ?? 2
= ?? 2
?? = ± ??
So, from the equations, there are two cases:
Case ?? :
?? = ??
In this case, since ?? and ?? are natural logarithms of positive numbers ?? and ?? respectively, this implies
that ?? = ?? . However, this cannot be true as a, ?? , and ?? are given to be distinct positive real numbers.
Case 2 :
?? = - ??
In this case, ln ? ?? = -ln ? ??
? ?? × ?? = 1
? ?? =
1
??
Also, ?? + ?? = 0 (from the equations above)
? ? ln ? ?? + ln ? ?? = 0
? ? ln ? ( ?? × ?? ) = 0
? ? ?? × ?? = 1
Given ?? =
1
?? and ?? × ?? = 1 ? ?? = ??
Thus, in the case where ?? = - ?? , the possible values are:
?? =
1
?? ?? = ??
If ???? = 1 ? ( 2 ?? )
ln ? ?? = 1
? ?? = 1 / 2
So, 6 ?? + 5 ???? = 6 (
1
2
) + 5 = 3 + 5 = 8
Q2: Let ?? = { ?? : ??????
?? ? ( ?? ?? ?? - ?? + ???? ) - ??????
?? ? (
?? ?? · ?? ?? ?? - ?? + ?? ) = ?? }. Then the maximum value of ?? for
which the equation ?? ?? - ?? ( ?
?? ? ?? ? ?? )
?? ?? + ?
?? ? ?? ? ( ?? + ?? )
?? ?? = ?? has real roots, is [JEE Main 2023
(Online) 25th January Morning Shift]
Ans: 25
Read MoreFAQs on Logarithm: JEE Main Previous Year Questions (2021-2026)
| 1. What are the main logarithm formulas I need for JEE Main exams? |  |
Ans. Essential JEE logarithm formulas include product rule (log_a(xy) = log_a x + log_a y), quotient rule (log_a(x/y) = log_a x - log_a y), power rule (log_a(x^n) = n·log_a x), and change of base formula (log_a x = log_b x / log_b a). Students should also memorise log_a 1 = 0 and log_a a = 1. These foundation formulas appear repeatedly across previous year JEE questions from 2021-2026.
| 2. How do I solve logarithmic equations that appear in JEE Main previous years? | |
Ans. To solve logarithmic equations, first identify the base and apply appropriate logarithm properties to simplify. Convert logarithmic form to exponential form when needed. Always verify solutions satisfy domain restrictions (arguments must be positive). JEE Main questions often combine multiple logarithm rules within single problems, so practise recognising when to use product, quotient, or power rules sequentially.
| 3. Why do logarithm questions in JEE combine with exponential functions so often? | |
Ans. Logarithms and exponentials are inverse operations-understanding their relationship deepens problem-solving ability. JEE Main tests this connection frequently because it assesses conceptual clarity beyond formula memorisation. Questions mixing both functions require students to convert between forms, apply transformations, and solve complex inequalities. This dual-function approach separates surface-level learners from deeper mathematical thinkers.
| 4. What's the most common mistake students make with logarithm domain in JEE questions? | |
Ans. The most frequent error is ignoring domain restrictions-forgetting that logarithm arguments must always be positive. Students often lose marks by finding mathematically correct solutions that violate this constraint. When solving log equations, always check that your final answer makes the original expression's argument greater than zero. Previous year JEE papers frequently include trap answers exploiting this oversight.
| 5. How should I use change of base formula to handle mixed-base logarithm problems in JEE? | |
Ans. Change of base formula converts any logarithm to a common or natural base: log_a x = ln x / ln a or log_a x = log₁₀ x / log₁₀ a. This technique simplifies expressions with multiple bases into manageable form. JEE Main previous year questions often present problems with bases like 2, 3, 5, or 7 that become tractable only after applying this conversion strategically.
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