$$ { \log_{5} }^2 x + \log_{5x} {5\over x } = 1 $$
My progress $$\begin{align} &{ \log_{5} }^2 x + \log_{5x} {5\over x } = 1 \\[2ex] \implies & { \log_{5} }^2 x + \log_{5x} {25} \cdot (5x)^{-1} = 1\\[2ex] \implies & { \log_{5} }^2 x + \log_{5x} {25} = 2 \\[2ex] \implies & { \log_{5} }^2 x + \frac{1}{ \log_{5^2} 5x } = 2\\[2ex] \implies & { \log_{5} }^2 x + \frac{1}{ \log_{5^2} 5 + \log_{5^2}x } = 2\\[2ex] \implies & { \log_{5} }^2 x + \frac{1}{ 2 + 2\log_{5} x } = 2\\[2ex] \end{align}$$ Taking $ { \log_{5} x } = y $
I get a cubic expression , Solving it I get $x = 5 , 1 , \Large \frac{1}{25} $
Is there a shorter alternative ?
Notice, $$(\log_5x)^2+\log_{5x}\left(\frac{5}{x}\right)=1$$ $$(\log_5x)^2+\left(\frac{\log_5\frac{5}{x}}{\log_5(5x)}\right)=1$$ $$(\log_5x)^2+\left(\frac{1-\log_5 x}{1+\log_5x}\right)=1$$ $$(\log_5x)^3+(\log_5x)^2+1-\log_5x=1+\log_5x$$ $$(\log_5x)^3+(\log_5x)^2-2\log_5x=0$$ let $\log_5x=t$ $$t^3+t^2-2t=0$$ $$t(t-1)(t+2)=0\implies t=0, 1, -2$$ $$\implies x=1, 5, \frac{1}{25}$$