Is it possible to clear the x using the Lambert function?

Question

$ y = \frac{x^2}{4} - \frac{ln(x)}{2} $

Solving, I get to:

$ e^{4y} = \frac{e^{x^2}}{x^2} $

But I don't know how to continue.

Answer 1

$$y=\frac{x^2}{4}-\frac{\ln(x)}{2}\Longleftrightarrow$$ $$y=\frac{x^2}{4}-\frac{2\ln(x)}{4}\Longleftrightarrow$$ $$y=\frac{x^2-2\ln(x)}{4}\Longleftrightarrow$$ $$4y=x^2-2\ln(x)\Longleftrightarrow$$ $$\exp[4y]=\exp\left[x^2-2\ln(x)\right]\Longleftrightarrow$$ $$\exp[4y]=\frac{\exp\left[x^2\right]}{x^2}\Longleftrightarrow$$

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Sustitute $x^2=u$

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$$\exp[4y]=\frac{\exp\left[u\right]}{u}\Longleftrightarrow$$ $$u=-\text{W}_n\left[-\frac{1}{e^{4y}}\right]\Longleftrightarrow$$ $$x^2=-\text{W}_n\left[-\frac{1}{e^{4y}}\right]\Longleftrightarrow$$

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Notice:

• $$\sqrt{\text{W}_n\left[-\frac{1}{e^{4y}}\right]}\ne0$$
• $$n\in\mathbb{Z}$$
• $\text{W}_k(z)$ is the analytic continuation of the product log function

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$$x=\pm\sqrt{-\text{W}_n\left[-\frac{1}{e^{4y}}\right]}$$