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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

 Category: JEE Advanced Important Questions Content-Type: Text, Images, Videos and PDF Exam: JEE Advanced Chapter Name: Logarithms Academic Session: 2024 Medium: English Medium Subject: Mathematics Available Material: Chapter-wise Important Questions with PDF

## Secure a Top Score in Your JEE Advanced Maths Logarithms Exam with Our Practice Papers

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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

Solution:

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 = ?

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)

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

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

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

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

Solution:

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

Solution:

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?

Solution:

$\text { Given } \ln \left(x^2+15 a^2\right)-\ln (a-2)=\ln \left(\dfrac{8 a x}{a-2}\right)\\ \text { In }\left[\dfrac{x^2+15 a^2}{a-2}\right]=\ln\left(\dfrac{8 a x}{a-2}\right) \\ \Rightarrow \left(\dfrac{x^2+15 a^2}{a-2}\right)=\left(\dfrac{8 a x}{a-2}\right) \\ \Rightarrow x^2+15 a^2=8 a x \\ \Rightarrow x^2+15 a^2-8 a x=0$ —(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) $log_{243} 32$

Solution:

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) $\log _{243} 32$

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