If (x > 0,) the ( 1 + \frac{\log\_{e^{2}}x}{1!} + \frac{\lef... | Mathematics | SolveeIt

Question

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

$x^2$

$\sqrt{x}$

Answer

$\sqrt{x}$

Explanation

Solution

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

Mathematics Question on limits and derivatives

Solution