UPSC Exam > UPSC Questions > If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then...
Start Learning for Free
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to?
Most Upvoted Answer
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is e...
Explanation:
1. Given Function:
- The given function is \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).
2. Substitute \( 2x_1 + x_2 \) for \( x \):
- Substitute \( 2x_1 + x_2 \) for \( x \) in the function.
- We get \( f(2x_1 + x_2) = \log\left(\frac{1+2x_1 + x_2}{1-2x_1 - x_2}\right) \).
3. Use the Properties of Logarithms:
- We can simplify the expression using the properties of logarithms.
- \( \log(a/b) = \log(a) - \log(b) \).
- Applying this property, we get \( f(2x_1 + x_2) = \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
4. Final Expression:
- Therefore, \( f(2x_1 + x_2) = \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
5. Conclusion:
- The expression for \( f(2x_1 + x_2) \) in terms of logarithms is \( \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
1. Given Function:
- The given function is \( f(x) = \log\left(\frac{1+x}{1-x}\right) \).
2. Substitute \( 2x_1 + x_2 \) for \( x \):
- Substitute \( 2x_1 + x_2 \) for \( x \) in the function.
- We get \( f(2x_1 + x_2) = \log\left(\frac{1+2x_1 + x_2}{1-2x_1 - x_2}\right) \).
3. Use the Properties of Logarithms:
- We can simplify the expression using the properties of logarithms.
- \( \log(a/b) = \log(a) - \log(b) \).
- Applying this property, we get \( f(2x_1 + x_2) = \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
4. Final Expression:
- Therefore, \( f(2x_1 + x_2) = \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
5. Conclusion:
- The expression for \( f(2x_1 + x_2) \) in terms of logarithms is \( \log(1+2x_1 + x_2) - \log(1-2x_1 - x_2) \).
Attention UPSC Students!
To make sure you are not studying endlessly, EduRev has designed UPSC study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in UPSC.
|
Explore Courses for UPSC exam
|
|
Similar UPSC Doubts
Top Courses for UPSCView all
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to?
Question Description
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to?.
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? for UPSC 2024 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? covers all topics & solutions for UPSC 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to?.
Solutions for If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? in English & in Hindi are available as part of our courses for UPSC.
Download more important topics, notes, lectures and mock test series for UPSC Exam by signing up for free.
Here you can find the meaning of If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? defined & explained in the simplest way possible. Besides giving the explanation of
If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to?, a detailed solution for If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? has been provided alongside types of If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? theory, EduRev gives you an
ample number of questions to practice If f(x)=log (1+x1−x)𝑓𝑥=log𝑒 1+𝑥1-𝑥, then f(2x1+x2)𝑓2𝑥1+𝑥2 is equal to? tests, examples and also practice UPSC tests.
|
|
Explore Courses for UPSC exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup