Question

The solution set of the equation ${\log }_{x}3{\log }_{3x}3={\log }_{9x}3$ is similar to which of the following options :

• A. ${\log }_{x}3{\log }_{3/x}3={\log }_{x/9}3$
• B. $({\log }_{3}x{)}^2=2$
• C. ${\log }_2\left(\frac{1}{2}{\log }_2{\log }_3{x}^2\right)=0$
• D. $\sqrt{2}{\log }_{x}3=1$

Answer 1

Answer: (A), (B)

The correct options are
A ${\log }_{x}3{\log }_{3/x}3={\log }_{x/9}3$
B $({\log }_{3}x{)}^2=2$

${\log }_{x}3{\log }_{3x}3={\log }_{9x}3x>0,3x>0>9x>0\&x,3x,9x\ne 1\frac{1}{{\log }_{3}x{\log }_{3}3x}=\frac{1}{{\log }_{3}9x}{\log }_{3}x(1+{\log }_{3}x)=2+{\log }_{3}x({\log }_{3}x{)}^2=2$

$x=\{3^{-\sqrt{2}},3^{\sqrt{2}}\}$


${\log }_{x}3{\log }_{3/x}3={\log }_{x/9}3{\log }_{3}x(1-{\log }_{3}x)={\log }_{3}x-2({\log }_{3}x{)}^2=2\Rightarrow x=3^{\pm \sqrt{2}}$

${\log }_2\left(\frac{1}{2}{\log }_2{\log }_3{x}^2\right)=0$
$\Rightarrow x=\pm 9$


$\sqrt{2}{\log }_{x}3=1\Rightarrow x=3^{-\sqrt{2}}$