Question

If $a,b>0,a,b\ne 1,c>0$, then ${\log }_{a}c=\frac{{\log }_{b}c}{{\log }_{b}a}=\left({\log }_{b}c\right)\left({\log }_{a}b\right)$

The solution set of ${\left({\log }_{x}5\right)}^2+{\log }_{5x}\frac{5}{x}=1$ is

• A. $(1,5)$
• B. $(1,\frac{1}{25})$
• C. $(5,\frac{1}{25})$
• D. $(1,5,\frac{1}{25},\frac{1}{125})$

Answer 1

Answer: (C)

$(5,\frac{1}{25})$

${\left({\log }_{x}5\right)}^2+{\log }_{5x}\left(\frac{5}{x}\right)=1$
above equation is valid when $x>0,x\ne 1$
$\Rightarrow {\left({\log }_{x}5\right)}^2+\frac{1-{\log }_{x}5}{1+{\log }_{x}5}=1$
$\Rightarrow {\log }_{x}5({\log }_{x}5-1)({\log }_{x}5+2)=0$
$\Rightarrow {\log }_{x}5=0,1,-2$
$\Rightarrow x=1,5,\frac{1}{25}$
Since, $x\ne 1$
therefore, $x=5,\frac{1}{25}$