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Question Number 153840 by liberty last updated on 11/Sep/21

$$\log {}_e\left(x\right)+\log {}_x\left(e\right)+\log {}_{\left(\frac{e}{x}\right)}\left(x\right)=\frac{5}{2}$$$$x=?$$

Answered by Rasheed.Sindhi last updated on 11/Sep/21

$$\log {}_e\left(x\right)+\log {}_x\left(e\right)+\log {}_{\left(\frac{e}{x}\right)}\left(x\right)=\frac{5}{2}$$$$\frac{{\log }_{\mathrm{e}}\mathrm{x}}{{\log }_{\mathrm{e}}\mathrm{e}}+\frac{{\log }_{\mathrm{e}}\mathrm{e}}{{\log }_{\mathrm{e}}\mathrm{x}}+\frac{{\log }_{\mathrm{e}}\mathrm{x}}{{\log }_{\mathrm{e}}\left(\frac{\mathrm{e}}{\mathrm{x}}\right)}=\frac{5}{2}$$$${\log }_{\mathrm{e}}\mathrm{x}+\frac{1}{{\log }_{\mathrm{e}}\mathrm{x}}+\frac{{\log }_{\mathrm{e}}\mathrm{x}}{{\log }_{\mathrm{e}}\mathrm{e}-{\log }_{\mathrm{e}}\mathrm{x}}=\frac{5}{2}$$$${\log }_{\mathrm{e}}\mathrm{x}+\frac{1}{{\log }_{\mathrm{e}}\mathrm{x}}+\frac{{\log }_{\mathrm{e}}\mathrm{x}}{1-{\log }_{\mathrm{e}}\mathrm{x}}=\frac{5}{2}$$$${\log }_{\mathrm{e}}\mathrm{x}=\mathrm{y}\left(\mathrm{say}\right)$$$$\mathrm{y}+\frac{1}{\mathrm{y}}+\frac{\mathrm{y}}{1-\mathrm{y}}=\frac{5}{2}$$$$\mathrm{y}\left(2\mathrm{y}(1-\mathrm{y})\right)+2(1-\mathrm{y})+{2\mathrm{y}}^2=5\mathrm{y}(1-\mathrm{y})$$$${2\mathrm{y}}^2-{2\mathrm{y}}^3+2-2\mathrm{y}+{2\mathrm{y}}^2-5\mathrm{y}+{5\mathrm{y}}^2=0$$$${2\mathrm{y}}^3-{9\mathrm{y}}^2+7\mathrm{y}-2=0$$$$....$$$$...$$

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