The value of log_e left( 1 + ax^2 + a^2 + fra | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We are given the expression $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$.First,factor the expression inside the logarithm:$1 + ax^2 + a^

Vedclass · Accepted Answer

$a \left( x^2 + \frac{1}{x^2} \right) - \frac{a^2}{2} \left( x^4 + \frac{1}{x^4} \right) + \frac{a^3}{3} \left( x^6 + \frac{1}{x^6} \right) - \dots$

Vedclass · Answer

(B) We are given the expression $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$.First,factor the expression inside the logarithm:$1 + ax^2 + a^2 + \frac{a}{x^2} = (1 + ax^2) + \frac{a}{x^2}(1 + ax^2) = (1 + ax^2)(1 + \frac{a}{x^2})$.Now,use the property $\log(mn) = \log m + \log n$:$\log_e \left( (1 + ax^2)(1 + \frac{a}{x^2}) \right) = \log_e(1 + ax^2) + \log_e(1 + \frac{a}{x^2})$.Using the logarithmic series expansion $\log_e(1 + y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \dots$:$= \left( ax^2 - \frac{(ax^2)^2}{2} + \frac{(ax^2)^3}{3} - \dots \right) + \left( \frac{a}{x^2} - \frac{(a/x^2)^2}{2} + \frac{(a/x^2)^3}{3} - \dots \right)$.Grouping the terms by powers of $a$:$= a(x^2 + \frac{1}{x^2}) - \frac{a^2}{2}(x^4 + \frac{1}{x^4}) + \frac{a^3}{3}(x^6 + \frac{1}{x^6}) - \dots$

The value of $\log_e \left( 1 + ax^2 + a^2 + \frac{a}{x^2} \right)$ is

Similar Questions

For $|x| < 1$,the coefficient of $x^3$ in the expansion of $\log(1+x+x^2)$ in ascending powers of $x$ is (in $/3$)

The sum to infinity of the given series $\frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$ is

$1+\frac{1}{3 \cdot 2^2}+\frac{1}{5 \cdot 2^4}+\frac{1}{7 \cdot 2^6}+\ldots$ is equal to

The expansion $\log_e(1 + x) = \sum\limits_{i = 1}^\infty \left[ \frac{(-1)^{i + 1}x^i}{i} \right]$ is defined for:

$1 + \frac{(\log_e n)^2}{2!} + \frac{(\log_e n)^4}{4!} + \dots = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module