![Expand ( [ log ( 1 + x ) ] ^ { 2 } ) is ascending prower of ( x ) , concilnunto fourth power of ( x ) ." Expand ( [ log ( 1 + x ) ] ^ { 2 } ) is ascending prower of ( x ) , concilnunto fourth power of ( x ) ."](https://toppr-doubts-media.s3.amazonaws.com/images/3816188/b0a35928-0b65-47e1-9aab-ad99487c5acc.jpg)
Expand ( [ log ( 1 + x ) ] ^ { 2 } ) is ascending prower of ( x ) , concilnunto fourth power of ( x ) ."
![If y = x^{3} log left ( dfrac{1}x{} right ), the prove that xdfrac{d^{2}y}{dx^{2}} - 2dfrac{dy}{dx} + 3x^{2} = 0. If y = x^{3} log left ( dfrac{1}x{} right ), the prove that xdfrac{d^{2}y}{dx^{2}} - 2dfrac{dy}{dx} + 3x^{2} = 0.](https://haygot.s3.amazonaws.com/questions/1697588_1782923_ans_b2e753a35f954a64a274f1c407d34c4b.jpg)
If y = x^{3} log left ( dfrac{1}x{} right ), the prove that xdfrac{d^{2}y}{dx^{2}} - 2dfrac{dy}{dx} + 3x^{2} = 0.
![algebra precalculus - Help with $\frac12 \log_2 x - \frac1{\log_2 x} = \frac76$ - Mathematics Stack Exchange algebra precalculus - Help with $\frac12 \log_2 x - \frac1{\log_2 x} = \frac76$ - Mathematics Stack Exchange](https://i.stack.imgur.com/U8Uem.jpg)
algebra precalculus - Help with $\frac12 \log_2 x - \frac1{\log_2 x} = \frac76$ - Mathematics Stack Exchange
![Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$ - Mathematics Stack Exchange Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$ - Mathematics Stack Exchange](https://i.stack.imgur.com/amoOG.png)
Intuition behind logarithm inequality: $1 - \frac1x \leq \log x \leq x-1$ - Mathematics Stack Exchange
Expand log (1 + e^x) in ascending powers of x up to the term containing x^4. - Sarthaks eConnect | Largest Online Education Community
![Equazioni esponenziali e logaritmiche: \frac{{ \log{{\left({x}-{1}\right)}}}}{ \log{{\left({x}^{3}-{8}{x}+{5}\right)}}}=\frac{1}{{3}} Equazioni esponenziali e logaritmiche: \frac{{ \log{{\left({x}-{1}\right)}}}}{ \log{{\left({x}^{3}-{8}{x}+{5}\right)}}}=\frac{1}{{3}}](https://www.skuola.net/news_foto/2017/10/equ_log_e1.jpg)