Playing with Logarithms

$\large (\log_5 x)^2 + \log_{5x} \left( \frac{5}{x}\right) = 1$ $\large \left(\dfrac{ln x }{ln 5}\right)^2 + \left( \dfrac{ln 5 - ln x }{ln 5 + ln x} \right) = 1$ Substitute $\ln x = t$ $\large \left(\dfrac{t }{ln 5}\right)^2 + \left( \dfrac{ln 5 - t }{ln 5 + t} \right) = 1$ Taking LCM and cross multiplying, $\large t(t^2 + t\ln 5 - 2(\ln 5)^2 = 0$ $\large t(t+2\ln 5)(t - \ln 5) = 0$ $t = 0 , \ln 5 , -2 \ln 5$ $\therefore x \in ( 1 , 5 , \frac{1}{25} )$

$The~equation~ is~ converted ~to~base ~5,\\ ~~~~\\ and ~we~get~(log_5 x)^2+ \dfrac{log_5 5-log_5 x}{ log_5 5+log_5 x} -1=0.\\ ~~~~\\ \implies~~( log_5 x)^2=\dfrac{2log_5 x}{1+ log_5 x}.\\ ~~~~\\ \therefore~~ log_5 x=0,~~\implies~x=1. \\ ~~~~\\ OR~~(log_5 x)^2*(1+log_5 x)=2log_5 x,~~~if~log_5 x \neq~0,\\ ~~~~\\ \implies~(log_5 x)(1+log_5 x)=2.\\ ~~~~\\ Solving~quadratic~~(log_5 x)^2+(log_5 x)-2=0,\\ ~~~~\\ log_5 x=\dfrac{-1\pm~\sqrt{1+8}}2. \\ ~~~~\\ \therefore~~log_5 x=1~~or~~log_5 x=-2.\\ ~~~~\\ \implies~roots,~~x=5,~~or~~x=5^{-2}.\\ ~~~~\\ Sum~of~roots~~=1+5+1/25=6.04$

$OR$

$The~equation~ is~ converted ~to~base ~x,~~and ~let~p=log_x 5.\\ ~~~~\\ Condition~for~converting~the~ base~is,~new~base~\neq~1.\\ ~~~~\\ On~checking~x=1~is~a~solution. \\ ~~~~\\ When~x\neq~1:-\\ ~~~~\\ log_5 x=\dfrac {log_x x}{log_x 5}~~if~~ log_x 5~\neq~0.\\ ~~~~\\ and ~the~equation~is~\dfrac 1 {p^2}+ \dfrac{p-1}{p+1} -1=0.\\ ~~~~\\ \implies~~\dfrac 1 {p^2}=\dfrac 2 {p+1}.\\ ~~~~\\ \therefore~quadratic~~2p^2 - p - 1=0.\\ ~~~~\\ \implies~p=1,~~ - \frac 1 2.\\ ~~~~\\ Gives~x=5,~~x=\frac 1 {25}. \\ ~~~~\\ On~checking~{\Large~~\color{#D61F06}{x=\frac 1 {25}, ~1,~5}}~~are~ solutions. \\ ~~~~\\ \therefore~\frac 1 {25}+1+5=6.04.$