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JEE Advanced Logarithms Important Questions

Last updated date: 23rd May 2024
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JEE Advanced Maths Logarithm Important Questions from PYQs with Solutions

Find here some of the most Important Questions Logarithms Jee Advanced to make your study process easier and more reliable for JEE Advanced Exams. These solutions are prepared by the best maths teachers in the country to help you perform exceedingly well in the actual exams. The logarithm JEE Advanced important questions are provided along with solutions allowing you to prepare perfectly for the upcoming exams. Scroll down and download the JEE Advanced Logarithms Important Questions given here as free PDF downloads.


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JEE Advanced Logarithms Important Questions

Logarithms in Maths is the inverse function to exponentiation. It has its applications in algebra as well as in trigonometric functions. Sums based on logarithms are very tricky. But with the proper understanding of the concepts and repeated practise, the sums here are quite scoring. The biggest advantage of this chapter is the fact that apart from being easy to solve, the answers here are easy to cross-check as well. 

Following are the few important topics that will be covered in this chapter. You can start your preparation on logarithms with these topics:

  • Remainder Theorem

  • Factor Theorem

  • Indices

  • Laws Of Indices

  • Intervals - Open Interval, Closed Interval, Open-closed Interval, Closed- Open Interval

  • Principal Properties Of Logarithm

  • Exponentials And Logarithmic Functions

  • Logarithmic Differentiation

  • Logarithmic Integration

Formulaic Support: Boosting Understanding with  JEE Advanced Logarithms Important Questions

If you're aiming for effective learning, consider employing a strategic approach that integrates Logarithms important questions for JEE Advanced with essential formulas and crucial questions. Formulas serve as concise explanations of fundamental concepts, while solved questions provide a comprehensive learning experience. A strong foundation is established through JEE Advanced Logarithms Important Questions, and formulas aid in comprehending concepts and addressing problems. This fusion streamlines complex topics and enhances memory retention, ultimately boosting your comprehension and academic performance.

Most Important Formulas of Logarithms for JEE Advanced

1. Basic Logarithmic Properties:

  • $\log_a(xy) = \log_a(x) + \log_a(y)$

  • $\log_a\left(\dfrac{x}{y}\right) = \log_a(x) - \log_a(y)$

  • $\log_a(x^n) = n\log_a(x)$

2. Change of Base Formula:

  • $\log_a(b) = \dfrac{\log_c(b)}{\log_c(a)}$

3. Common Logarithms:

  • $\log_{10}(x) = \log(x)$

  • $\log_{10}(10^x) = x$

4. Natural Logarithms:

  • $\ln(x) = \log_e(x)$

5. Logarithmic Identities:

  • $\log_a(1) = 0$

  • $\log_a(a) = 1$

  • $\log_a(a^n) = n$

  • $a^{\log_a(x)} = x$

  • $\log_a(xy) = \log_a(x) + \log_a(y)$

Logarithms JEE Advanced Important Questions

The Logarithms Important Questions JEE Advanced are one of the most resourceful study materials. These questions and their respective study materials are given after a thorough analysis of the necessary syllabus. Also, numerous previous year's question papers are taken into consideration while preparing these questions. The solutions are filled with all the tricks and various shortcut techniques allowing you to solve all the problems with utmost accuracy and speed.

Solved JEE Advanced Important Questions of Logarithms

1. Let $(x_0, y_0)$ be the solution of the following equations then, $(2 x)^{\ln 2}=(3 y)^{\ln 3}$ and $3^{\ln x}=2^{\ln y}$. Then $x_0$ is


Given, $(2 x)^{\ln 2}=(3 y)^{\ln 3}$ —(1)

Taking log on both sides

$\log 2 \log (2 x)=\log 3 \log (3 y) \\ \log 2(\log 2+\log x)=\log 3(\log 3+\log y)$—(2)

Also given, $3^{\ln x}=2^{\ln y}$ —(3)

Taking log on both sides

$\log x \log 3=\log y \log 2 \\ \log y=\log x \log 3 / \log 2$ —(4)

Substituting equation (4) in (2), we get

$\log 2(\log 2+\log x)=\log 3(\log 3+\log x \log 3 / \log 2) \\ (\log 2)^2+\log 2 \log x=(\log 3)^2+\log x(\log 3)^2 / \log 2 \\ {\left[\log 2-(\log 3)^2 / \log 2\right] \log x=(\log 3)^2-(\log 2)^2} \\ \log x=\left[(\log 3)^2-(\log 2)^2\right] /\left[\left((\log 2)^2-(\log 3)^2\right) / \log 2\right] \\ \log x=-\log 2 \\ \log x = log 2^{-1} \\ x = 2^{-1} \\ x = \dfrac{1}{2}$

2. If 3x = 4x-1, then x = ?


Given that 3x = 4x-1

Taking log on both sides, we get

log 3x = log 4x-1

x log 3 = (x-1) log 4

x log 3 = x log 4 – log 4

log 4 = x(log 4 – log 3)

$x = \dfrac{\log 4}{(\log 4 - \log 3)}$

$x = \dfrac{\log 4}{1 - \dfrac{\log 3}{\log 4}}$

$x = \dfrac{1}{1 - \log_4 3}$

Hence the value of $x$ is $\dfrac{1}{1 - log_4 3}$.

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


Given, $x+\log _{10}\left(2^x+1\right)=x \log _{10} 5+\log _{10} 6$

$x\left[1-\log _{10} 5\right]+\log _{10}\left(2^x+1\right)=\log _{10} 6 \\ x\left[\log _{10} 10-\log _{10} 5\right]+\log _{10}\left(2^x+1\right)=\log _{10} 6 \\ x \log _{10} 2+\log _{10}\left(2^x+1\right)=\log _{10} 6 \\ \log _{10} 2^x\left(2^x+1\right)=\log _{10} 6 \\ \left(2^x\right)^2+2^x-6=0 \\ 2^x=2 \text { or } 2^x=-3$

Hence $x = 1$ is the number of positive integers

4. If $\log _7 2=k$, then $\log _{49} 28$ is equal to


Given $\log _7 2=k$

$\log _{49} 28= \\ \log _{7^2} 28 \\ \dfrac{1}{2} \log _7 28 \\ \dfrac{1}{2} \log _7(4 \times 7) \\ \dfrac{1}{2} \log _7 4 + \dfrac{1}{2} \log _7 7 \\ \dfrac{1}{2} \log _7 2^2 + \dfrac{1}{2} \\ \log _7 \dfrac{2+1}{2} \\ \dfrac{k + 1}{2} \\ \dfrac{2K+1}{2}$

5. If x = 9 is a solution of $\ln \left(x^2+15 a^2\right)-\ln (a-2)=\ln (\dfrac{8 a x}{(a-2)})$ then a and x is equal to?


$ \text { Given } \ln \left(x^2+15 a^2\right)-\ln (a-2)=\ln \left(\dfrac{8 a x}{a-2}\right)\\ \text { In }\left[\dfrac{x^2+15 a^2}{a-2}\right]=\ln\left(\dfrac{8 a x}{a-2}\right) \\ \Rightarrow \left(\dfrac{x^2+15 a^2}{a-2}\right)=\left(\dfrac{8 a x}{a-2}\right) \\ \Rightarrow x^2+15 a^2=8 a x \\ \Rightarrow x^2+15 a^2-8 a x=0$ —(1)

Now, Given $x=9$ is a root.

$\Rightarrow 81+15 a^2-72 a=0 \\ \Rightarrow 5 a^2-24 a+27=0 \\ \Rightarrow(5 a-9)(a-3)=0 \\ \Rightarrow a = \dfrac{9}{5} \text { or } a=3$

Put value of $a$ in (i)

When $a = \dfrac{9}{5}$, we get $x=9$ or $x = \dfrac{27}{5}$

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

6. If $\log_{10} 5 = a$ and $\log_{10} 3 = b$ then find 

(a) $\log_{30} 8$

(b) $log_{40} 15$

(c) $log_{243} 32$


Given $\log _{10} 5=a$ and $\log _{10} 3=b$

(a) For $\log _{30} 8$

$\log _{30} 8= \dfrac{\log 2^3 }{\log (3 \times 10)} \\ \Rightarrow \dfrac{3 \log 2}{(\log 3+\log 10)} \\ \Rightarrow \dfrac{3 \log (10 / 5)}{(1+\log 3)} \\ \Rightarrow \dfrac{3(1-\log 5)}{(1+\log 3)} \\ \Rightarrow \dfrac{3(1-a)}{(1+b)}$

Hence $\log _{30} 8 = \dfrac{\log 2^3 }{\log (3 \times 10)} = \dfrac{3(1-a)}{(1+b)}$

(b) $\log _{40} 15$

$\log _{40} 15 = \dfrac{\log 15}{\log 40} \\ = \dfrac{\log (3 \times 5)}{\log (10 \times 4)} \\ = \dfrac{(\log 3+\log 5)}{\left(1+\log 2^2\right)} \\ \dfrac{(\log 3+\log 5)}{(1+2 \log 2)} \\ \dfrac{(\log 3+\log 5)}{(1+2 \log (10 / 5))} \\ \dfrac{(\log 3+\log 5)}{(1+2(1-\log 5))} \\ \dfrac{(\log 3+\log 5)}{(1+2-2 \log 5)} \\ \dfrac{(b+a)}{(3-2 a)}$

Hence $\log _{40} 15 = \dfrac{(b+a)}{(3-2 a)}$

(c) $\log _{243} 32$

$\log _{243} 32= \dfrac{\log 32}{\log 243} \\ = \dfrac{\log 2^5}{\log 3^5} \\ \dfrac{5 \log 2}{5 \log 3} \\ \dfrac{\log 2}{\log 3} \\ \dfrac{(1-\log 5)}{\log 3}$

(Since, $\log 2 = \log \dfrac{10}{5} = \log _{243} 32 = \log 10-\log 5=1-\log 5) 

Then, $\log _{243} 32 = \dfrac{(1-a)}{b}$

7. The value of $\left(\left(\log _2 9\right)^2\right)^{\dfrac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\dfrac{1}{\log _4 7}}$ is?


Given, $\left(\left(\log _2 9\right)^2\right)^{\dfrac{1}{\log _2\left(\log _2 9\right)}} \times(\sqrt{7})^{\dfrac{1}{\log _4 7}}$

$\log _2 9^{\dfrac{2}{\log _2\left(\log _2 9\right)}} \times 7^{\dfrac{1 / 2}{\log _4 7}} \\ =\left(\log _2 9\right)^{2 \log _{\log _2 9}^2} \times 7^{\dfrac{1}{2} \log _7 4} \\ =4 \times 2=8$


8. If $A = \log _2 \log _2 \log _4 256+\log _{\sqrt{2 }} 2$, then A is equal to


We know that, $\log _a(m)^n=n \log _a m$

$A = \log _2 \log _2 \log _4 256+\log _{\sqrt{2}} 2 \\ A =\log _2 \log _2 \log _4(4)^4+\log _{\sqrt{2 }} \sqrt{ 2} ^2 \\ A=\log _2 \log _2 4+2(2) \\ A=\log _2 \log _2 2^2+4 \\ =\log _2 2+4 \\ A=1+4\left(\log _a a=1\right) \\ A=5$

Next Steps: Further Resources for Important Questions of JEE Advanced 2024 Logarithms

Explore additional resources beyond JEE Advanced 2024 Logarithms Important Questions. Students need to consider extra materials like practice papers, mock tests, PYQPs, etc., alongside important Logarithms questions to gain a well-rounded preparation, refine their exam strategies, and build confidence for the JEE Advanced exam.

Ready to take your JEE Advanced 2024 Logarithms preparation up a notch? Delve into these crucial resources:

Other Important Links For JEE Advanced 2024

JEE Advanced Logarithms Practice Papers 2024

JEE Advanced Complex Numbers Mock Test 2024

JEE Advanced Logarithms Mock Test 2024

JEE Advanced Logarithms Revision Notes

Maths Important Questions for JEE Advanced 2024 Chapters: Links Available

Make your JEE Advanced Maths studying for 2024 even better using our Important Questions organised by chapter. Find the links in the table below for easy access.

JEE Advanced 2024: Chapter-Wise Links to Maths Important Questions PDF


Chapter Name


Sets, Relations and Functions






Probability and Statistics 




Analytical Geometry 


Differential Calculus 


Integral Calculus 




Quadratic Equations


Mastering logarithms is crucial for success in the JEE Advanced exam. To excel, focus on understanding the basic properties and rules of logarithms, such as the product, quotient, and power rules. Additionally, practice solving a variety of logarithmic equations and inequalities to strengthen your problem-solving skills. Pay special attention to questions involving logarithmic identities and their applications in calculus and algebra. Consistent practice with challenging problems will help solidify your understanding and boost your confidence for the exam. Remember, a strong grasp of logarithms can significantly enhance your overall performance in the JEE Advanced.

FAQs on JEE Advanced Logarithms Important Questions

1. How to download JEE Advanced Logarithms Important Questions?

JEE Advanced Logarithms Important Questions are a set of questions especially designed for the JEE Advanced students to build a better understanding of various concepts introduced in the logarithms chapter. Logarithms is a chapter with different new concepts which might not be known to some of the students. You don’t need to worry about downloading the JEE Advanced Logarithms Important Questions as you can easily find them at the Vedantu website. You just need to log in to the website and download whatever you desire.

2. Why choose Vedantu for downloading the JEE Advanced Logarithms Important Questions?

JEE Advanced Logarithms Important Questions require the utmost attention and intelligence of the students for maintaining a good rank. Vedantu is one of the most trustworthy websites for different reasons. It keeps the solutions very authentic and unique which is keenly designed by the expert teachers at Vedantu. The experts are designing the questions with utmost care and keeping in mind the latest JEE Advanced syllabus. Hence, Vedantu can be completely trusted from the point of view of the students as they are benefiting from the solutions at Vedantu.

3. What are the Benefits of Solving JEE Advanced Logarithms Important Questions?

The following are some of the key benefits of the JEE Advanced Logarithms Important Questions:

  • Solved JEE Advanced Logarithms Important Question Papers are very helpful for students to understand each Mathematics concept.

  • Get a good idea of ​​the type of questions asked in each topic in the large JEE Advanced Maths Exam.

  • It also helps aspirants to gain an in-depth knowledge of the JEE Main pattern of the Maths paper.

  • By preparing with these JEE Mathematics Solved papers, students can assess their preparation levels.

  • One of the best benefits students can get from this excellent preparation tool is getting more marks in the exam.

4. Why Are Previous Years Questions Important to Prepare for the JEE Advanced Maths Paper?

Selection of key questions and sound solutions prepared by subject specialists is necessary during the preparation of the JEE Advanced Maths Exam Paper as it helps you to get more information on additional marks in the exam. Solutions are helpful when you quickly finish your homework and prepare for exams. All questions and answers from the IIT JEE (Enhanced) Previous-Year Questionnaire for Chapter Logarithms Statistics and its structures are provided in Vedantu in order. Vedantu will assist you online with any doubts/clarifications.

5. How can a student get good marks in advanced JEE tests in Maths?

The mantra of success for JEE Advanced is practice and hard work. Gone are the days when students spent hours trying out a single question. Now is the time for many questions to choose from. JEE Mathematics questions test a learner's knowledge and ability. We have excellent notes prepared by experts at Vedantu to better match the exam requirement. Focus is given to problem-solving skills and small tips and tricks to make it faster and easier. We, the experts at Vedantu will make sure that the students understand the concept well.

6. What are the rules when adding logs?

You can only add logs if they have the same base. When adding logs with the same base, you multiply the numbers inside the logs (arguments).

7. When adding logs do you multiply?

Yes, adding logs with the same base is done by multiplying the arguments.

8. What are the 7 rules of logarithms?

There are several important rules of logarithms, but common ones include:

  1. Product Rule: $\log_b(mn) = \log_b(m) + \log_b(n)$ (log of a product equals the sum of logs of factors)

  2. Quotient Rule: $\log_b \left(\dfrac{m}{n}\right) = \log_b(m) - \log_b(n)$ (log of a quotient equals log of numerator minus log of denominator)

  3. Power Rule: $\log_b(m^n) = n  \log_b(m)$ (log of an exponent equals exponent times log of base)

9. How do you cancel logarithms?

You can only cancel logarithms if they have the same argument (the number inside the log) AND the same base.

10. When two logs are divided?

Dividing logs with the same base is the same as subtracting their arguments. So, $\dfrac{\log_b(m)}{\log_b(n)}$ is equal to $\log_b(m) - \log_b(n)$.

11. How do you expand logs?

Logarithms can sometimes be expanded using the product and quotient rules. For instance, log(xy) can be expanded to log(x) + log(y).

12. Who invented logarithm?

John Napier, a Scottish mathematician, is generally credited with inventing logarithms in the early 17th century.

13. Why is logarithm used?

Logarithms are useful in many fields because they can compress large numbers into smaller ones and simplify complex operations involving exponents. They have applications in physics, computer science, engineering, and various other areas.

14. What are the 3 types of logarithms?

There are three main types of logarithms:

  1. Common Logarithms (base-10): Represented as log(x), where x is any positive number.

  2. Natural Logarithms (base-e): Represented as ln(x), where e is a mathematical constant (approximately 2.71828).

  3. Logarithms with Any Base (b): Represented as $\log_b(x)$, where b is any positive number and b ≠ 1.