Question

If ${\log }_{2x}2\cdot {\log }_{x}2={\log }_{4x}2,$ then number of integral values of $x$ is equal to

• A. $2$
• B. $4$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

${\log }_{2x}2\cdot {\log }_{x}2={\log }_{4x}2$
For $\log$ to be defined
$2x>0$ and $2x\ne 1...\left(1\right)$
$x>0$ and $x\ne 1...\left(2\right)$
$4x>0$ and $4x\ne 1...\left(3\right)$
From equations $\left(1\right),\left(2\right)$ and $\left(3\right)$
$x>0$ and $x\ne \frac{1}{4},\frac{1}{2},1$

Now,
${\log }_{2x}2\cdot {\log }_{x}2={\log }_{4x}2$
$\Rightarrow \frac{1}{{\log }_{2}2x}\cdot \frac{1}{{\log }_{2}x}=\frac{1}{{\log }_{2}4x}$
$\Rightarrow {\log }_{2}2x\cdot {\log }_{2}x={\log }_{2}4x$
$\Rightarrow (1+{\log }_{2}x){\log }_{2}x=2+{\log }_{2}x$
$\Rightarrow ({\log }_{2}x{)}^2=2$
$\Rightarrow {\log }_{2}x=\pm \sqrt{2}$
$\Rightarrow x=2^{\pm \sqrt{2}}\notin Z$

Hence, the number of integral values of $x$ is $0.$