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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} $

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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.
answered Apr 16, 2014 by sreemathi.v
 

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