log-x-2-log2-x-e-2-1-e-x-

Question Number 161215 by mathlove last updated on 14/Dec/21

\log \underset{2}{x} + l o g \underset{x}{2} = \frac{e^{2} + 1}{e} x = ?

Commented by cortano last updated on 14/Dec/21

l e t {\begin{cases} \log_{2} (x) = u \\ \log_{x} (2) = \frac{1}{u} \end{cases}

\Rightarrow u + \frac{1}{u} = e + e^{- 1}

\Rightarrow u^{2} - (e + e^{- 1}) u + 1 = 0

\Rightarrow u = \frac{e + e^{- 1} \pm \sqrt{e^{2} + e^{- 2} - 2}}{2}

\Rightarrow u = \frac{e + e^{- 1} \pm \sqrt{{(e - e^{- 1})}^{2}}}{2}

\Rightarrow {\begin{cases} u_{1} = e \Rightarrow \log_{2} (x) = e \Rightarrow x = 2^{e} \\ u_{2} = e^{- 1} \Rightarrow \log_{2} (x) = e^{- 1} \Rightarrow x = 2^{\frac{1}{e}} \end{cases}

Answered by Rasheed.Sindhi last updated on 14/Dec/21

\log_{2} x + \log_{x} 2 = \frac{e^{2} + 1}{e}

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

\log_{2} x = e, \frac{1}{e} \Rightarrow x = 2^{e}, 2^{1 / e}

Commented by cortano last updated on 14/Dec/21

h o w a b o u t {\begin{cases} \log_{2} (x) = \frac{1}{e} \\ \log_{x} (2) = e \end{cases} ?

Commented by Rasheed.Sindhi last updated on 14/Dec/21

{Y o u}^{'} r e r i g h t! I e x t e n d e d m y a n s w e r .