Discover the best answers at Westonci.ca, where experts share their insights and knowledge with you. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
[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]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.