Question

If $x > 0,$ the $ 1 + \frac{\log_{e^{2}}x}{1!} + \frac{\left(\log_{e^{2}}x\right)^{2}}{2!} + ....=$

• A. x
• B. $x^2$
• C. 2x
• D. $\sqrt{x}$

Answer 1

The Correct Option is D

$1+ \frac{\log_{e^{2}}x}{1!} + \frac{\left(\log_{e^{2}} x\right)^{2}}{2!} + .... $
$=e^{\log}e^{2} {^{x} }= e^{\frac{1}{2} \log_{e}x} = e^{\log_{e} \sqrt{x}} = \sqrt{x}$