Answered

At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Connect with a community of experts ready to provide precise solutions to your questions quickly and accurately. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

Lnx+Ln(2x)=1/e
Can someone help me solve this problem?

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]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.