Question

Value of $x$ that satisfies the equation $\left({\log }_{3}x\right)\left({\log }_{5}9\right)-{\log }_{x}25+{\log }_{3}2={\log }_{3}54$ is

• A. $25$
• B. $5$
• C. $\frac{1}{\sqrt{5}}$
• D. $\frac{1}{5}$

Answer 1

Answer: (A), (C)

$25$
$\frac{1}{\sqrt{5}}$

$\left({\log }_{3}x\right)\left({\log }_{5}9\right)-{\log }_{x}25+{\log }_{3}2={\log }_{3}54$
$\Rightarrow \left({\log }_{3}x\right)\left({\log }_{5}9\right)-{\log }_{x}25={\log }_{3}54-{\log }_{3}2$
$\Rightarrow \left({\log }_{3}x\right)\left({\log }_{5}9\right)-2{\log }_{x}5={\log }_{3}27[\because \log a-\log b=\log \frac{a}{b}]$
$\Rightarrow \left(\frac{{\log }_{e}x}{{\log }_{e}3}\right)\left(\frac{{\log }_{e}9}{{\log }_{e}5}\right)-2\frac{{\log }_{e}5}{{\log }_{e}x}=3[\because {\log }_{b}a=\frac{\log a}{\log b}]$
$\Rightarrow 2\left(\frac{{\log }_{e}x}{{\log }_{e}5}\right)-2\frac{{\log }_{e}5}{{\log }_{e}x}=3$
$\Rightarrow 2{\log }_{5}x-\frac{2}{{\log }_{5}x}=3$
Substitute ${\log }_{5}x=t$
$\Rightarrow 2t-\frac{2}{t}=3$
$\Rightarrow 2t^2-3t-2=0$
$\Rightarrow (2t+1)(t-2)=0$
$\Rightarrow t=-\frac{1}{2},t=2$
$\Rightarrow {\log }_{5}x=-\frac{1}{2},{\log }_{5}x=2$
$\Rightarrow 5^{-1/2}=x;5^2=x$
From the given eqn , it follows that $x>0,x\ne 1$
Hence, $x=\frac{1}{\sqrt{5}},25$ are the solutions of given eqn