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Question Number 200474 by Rupesh123 last updated on 19/Nov/23

Answered by witcher3 last updated on 19/Nov/23

$$\mathrm{erf}\left(\mathrm{x}\right)=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{-{\mathrm{t}}^2}\mathrm{dt}$$$$\ln (\mathrm{x}+\ln \left(\mathrm{x}\right))={\int }_0^5{\mathrm{e}}^{-{\mathrm{t}}^2}=\frac{\sqrt{\pi }}{2}\mathrm{erf}\left(5\right)$$$$\mathrm{x}+\ln \left(\mathrm{x}\right)=\ln \left({\mathrm{xe}}^{\mathrm{x}}\right)=\frac{\sqrt{\pi }}{2}\mathrm{erf}\left(5\right)$$$${\mathrm{xe}}^{\mathrm{x}}={\mathrm{e}}^{\frac{\sqrt{\pi }}{2}\mathrm{erf}\left(5\right)}$$$$\mathrm{x}=\mathrm{W}\left({\mathrm{e}}^{\frac{\sqrt{\pi }}{2}\mathrm{erf}\left(5\right)}\right)$$$$$$

Commented by Rupesh123 last updated on 20/Nov/23

Very elegant!

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