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

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, $(2x{)}^{\ln 2}=(3y{)}^{\ln 3}$ and $3^{\ln x}=2^{\ln y}$. Then $x_0$ is

Solution:

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

Taking log on both sides

$$\begin{gathered} \log 2\log (2x)=\log 3\log (3y) \\ \log 2(\log 2+\log x)=\log 3(\log 3+\log y) \end{gathered}$$—(2)

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

Taking log on both sides

$$\begin{gathered} \log x\log 3=\log y\log 2 \\ \log y=\log x\log 3/\log 2 \end{gathered}$$ —(4)

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

$$\begin{gathered} \log 2(\log 2+\log x)=\log 3(\log 3+\log x\log 3/\log 2) \\ (\log 2{)}^2+\log 2\log x=(\log 3{)}^2+\log x(\log 3{)}^2/\log 2 \\ [\log 2-(\log 3{)}^2/\log 2]\log x=(\log 3{)}^2-(\log 2{)}^2 \\ \log x=[(\log 3{)}^2-(\log 2{)}^2]/[((\log 2{)}^2-(\log 3{)}^2)/\log 2] \\ \log x=-\log 2 \\ \log x=log{2}^{-1} \\ x=2^{-1} \\ x={\displaystyle \frac{1}{2}} \end{gathered}$$

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

Solution:

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)

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

$x={\displaystyle \frac{\log 4}{1-{\displaystyle \frac{\log 3}{\log 4}}}}$

$x={\displaystyle \frac{1}{1-{\log }_{4}3}}$

Hence the value of $x$ is ${\displaystyle \frac{1}{1-lo{g}_{4}3}}$.

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

Solution:

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

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

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

4. If ${\log }_{7}2=k$, then ${\log }_{49}28$ is equal to

Solution:

Given ${\log }_{7}2=k$

$$\begin{gathered} {\log }_{49}28= \\ {\log }_{7^2}28 \\ {\displaystyle \frac{1}{2}}{\log }_{7}28 \\ {\displaystyle \frac{1}{2}}{\log }_7(4\times 7) \\ {\displaystyle \frac{1}{2}}{\log }_{7}4+{\displaystyle \frac{1}{2}}{\log }_{7}7 \\ {\displaystyle \frac{1}{2}}{\log }_7{2}^2+{\displaystyle \frac{1}{2}} \\ {\log }_7{\displaystyle \frac{2+1}{2}} \\ {\displaystyle \frac{k+1}{2}} \\ {\displaystyle \frac{2K+1}{2}} \end{gathered}$$

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

Solution:

$$\begin{gathered} \text{ Given }\ln (x^2+15a^2)-\ln (a-2)=\ln \left({\displaystyle \frac{8ax}{a-2}}\right) \\ \text{ In }\left[{\displaystyle \frac{x^2+15a^2}{a-2}}\right]=\ln \left({\displaystyle \frac{8ax}{a-2}}\right) \\ \Rightarrow \left({\displaystyle \frac{x^2+15a^2}{a-2}}\right)=\left({\displaystyle \frac{8ax}{a-2}}\right) \\ \Rightarrow x^2+15a^2=8ax \\ \Rightarrow x^2+15a^2-8ax=0 \end{gathered}$$ —(1)

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

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

Put value of $a$ in (i)

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

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) $lo{g}_{40}15$

(c) $lo{g}_{243}32$

Solution:

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

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

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

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

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

$$\begin{gathered} {\log }_{40}15={\displaystyle \frac{\log 15}{\log 40}} \\ ={\displaystyle \frac{\log (3\times 5)}{\log (10\times 4)}} \\ ={\displaystyle \frac{(\log 3+\log 5)}{(1+\log 2^2)}} \\ {\displaystyle \frac{(\log 3+\log 5)}{(1+2\log 2)}} \\ {\displaystyle \frac{(\log 3+\log 5)}{(1+2\log (10/5))}} \\ {\displaystyle \frac{(\log 3+\log 5)}{(1+2(1-\log 5))}} \\ {\displaystyle \frac{(\log 3+\log 5)}{(1+2-2\log 5)}} \\ {\displaystyle \frac{(b+a)}{(3-2a)}} \end{gathered}$$

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

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

$$\begin{gathered} {\log }_{243}32={\displaystyle \frac{\log 32}{\log 243}} \\ ={\displaystyle \frac{\log 2^5}{\log 3^5}} \\ {\displaystyle \frac{5\log 2}{5\log 3}} \\ {\displaystyle \frac{\log 2}{\log 3}} \\ {\displaystyle \frac{(1-\log 5)}{\log 3}} \end{gathered}$$

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

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

7. The value of ${\left({({\log }_{2}9)}^2\right)}^{{\displaystyle \frac{1}{{\log }_2({\log }_{2}9)}}}\times (\sqrt{7}{)}^{{\displaystyle \frac{1}{{\log }_{4}7}}}$ is?

Solution:

Given, ${\left({({\log }_{2}9)}^2\right)}^{{\displaystyle \frac{1}{{\log }_2({\log }_{2}9)}}}\times (\sqrt{7}{)}^{{\displaystyle \frac{1}{{\log }_{4}7}}}$

$$\begin{gathered} {\log }_2{9}^{{\displaystyle \frac{2}{{\log }_2({\log }_{2}9)}}}\times 7^{{\displaystyle \frac{1/2}{{\log }_{4}7}}} \\ ={({\log }_{2}9)}^{2{\log }_{{\log }_{2}9}^2}\times 7^{{\displaystyle \frac{1}{2}}{\log }_{7}4} \\ =4\times 2=8 \end{gathered}$$

8. If $A={\log }_2{\log }_2{\log }_{4}256+{\log }_{\sqrt{2}}2$, then A is equal to

Solution:

We know that, ${\log }_a(m{)}^n=n{\log }_{a}m$

$$\begin{gathered} A={\log }_2{\log }_2{\log }_{4}256+{\log }_{\sqrt{2}}2 \\ A={\log }_2{\log }_2{\log }_4(4{)}^4+{\log }_{\sqrt{2}}{\sqrt{2}}^2 \\ A={\log }_2{\log }_{2}4+2(2) \\ A={\log }_2{\log }_2{2}^2+4 \\ ={\log }_{2}2+4 \\ A=1+4({\log }_{a}a=1) \\ A=5 \end{gathered}$$

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:

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.

Conclusion

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.