# Sum of all integral roots of $(\log_5x)^2+\log_{5x}(\large\frac{5}{x})$$=1 is \begin{array}{1 1}(A)\;5&(B)\;25\\(C)\;1&(D)\;6\end{array} ## 1 Answer Clearly x\neq 0\;\;x > 0 x\neq \large\frac{1}{5} \log_{5x}(\large\frac{5}{x})=\large\frac{\log_55-\log_5x}{\log_55+\log_5x} \Rightarrow \large\frac{1-\log_5x}{1+\log_5x} Taking \log_5x=t,equation in question becomes t^2+\large\frac{1-t}{1+t}$$=1$
$\Rightarrow t^3+t^2+1-t=1+t$
$\Rightarrow t^3+t^2-2t=0$
$\Rightarrow t(t-1)(t+2)=0$
$\Rightarrow t=0,1,-2$
$\Rightarrow x=1,5,\large\frac{1}{25}$
Sum of integral roots =6
Hence (D) is the correct answer.