log-e-x-log-x-e-log-e-x-x-5-2-x-

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

\log_{e} (x) + \log_{x} (e) + \log_{(\frac{e}{x})} (x) = \frac{5}{2}

x = ?

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

\log_{e} (x) + \log_{x} (e) + \log_{(\frac{e}{x})} (x) = \frac{5}{2}

\frac{\log_{e} x}{\log_{e} e} + \frac{\log_{e} e}{\log_{e} x} + \frac{\log_{e} x}{\log_{e} (\frac{e}{x})} = \frac{5}{2}

\log_{e} x + \frac{1}{\log_{e} x} + \frac{\log_{e} x}{\log_{e} e - \log_{e} x} = \frac{5}{2}

\log_{e} x + \frac{1}{\log_{e} x} + \frac{\log_{e} x}{1 - \log_{e} x} = \frac{5}{2}

\log_{e} x = y (say)

y + \frac{1}{y} + \frac{y}{1 - y} = \frac{5}{2}

y (2 y (1 - y)) + 2 (1 - y) + {2 y}^{2} = 5 y (1 - y)

{2 y}^{2} - {2 y}^{3} + 2 - 2 y + {2 y}^{2} - 5 y + {5 y}^{2} = 0

{2 y}^{3} - {9 y}^{2} + 7 y - 2 = 0

\dots .

\dots