Answer :
[tex]D:x > 0\\\\ln(x)+ln(2x)=\frac{1}{e}\ \ \ \ |use\ ln(a)+ln(b)=ln(ab)\\\\ln(x\cdot2x)=\frac{1}{e}\\\\ln(2x^2)=\frac{1}{e}\ \ \ \ |use\ b=ln(e^b)\\\\ln(2x^2)=ln\left(e^\frac{1}{e}\right)\iff2x^2=e^\frac{1}{e}\ \ \ \ |divide\ both\ sides\ by\ 2\\\\x^2=\frac{e^\frac{1}{e}}{2}\iff x=\sqrt\frac{e^\frac{1}{e}}{2}\\\\x=\frac{\sqrt{e^\frac{1}{e}}}{\sqrt2}\cdot\frac{\sqrt2}{\sqrt2}\\\\\boxed{x=\frac{\sqrt{2e^\frac{1}{2}}}{2}}[/tex]
