Question

The positive integral solution of the equation ${\log }_x\sqrt{5}+{\log }_{x}5x=\frac{9}{4}+{\log }_x^2\sqrt{5}$ is

• A. Composite number
• B. Prime number
• C. Even number
• D. divisible by 3

Answer 1

The correct option is A Composite number

${\log }_x\sqrt{5}+{\log }_{x}5x=\frac{9}{4}+{\log }_x^2\sqrt{5}$

$=\frac{1}{2}{\log }_{x}5+{\log }_{x}5x=\frac{9}{4}+\frac{1}{4}+{\log }_x^{2}5$

$=\frac{1}{2}\left({\log }_{x}5+1\right)+{\log }_{x}5x=\frac{9}{4}+\frac{1}{4}+{\log }_x^{2}5$=

$=2{\log }_{x}5+4\left({\log }_{x}5\right)+4=9+\left({\log }_x^{2}5\right)$

$=\left({\log }_x^{2}5\right)-6\left({\log }_{x}5\right)+5=0$

$=({\log }_{x}5-1)({\log }_x^5-5)=0$

Then, $\Rightarrow {\log }_{x}5=1$

$\Rightarrow 5=x$

$\Rightarrow x=5$

And, $\Rightarrow {\log }_{x}5=5$

$\Rightarrow 5=x^5$

$\Rightarrow x=(5{)}^{1/5}$

So, $5$ is a composite number.