Question

The value of $x$ satisfying ${\log }_{16}x+{\log }_{x}16={\log }_{512}x+{\log }_{x}512$ is/are

• A. $\frac{1}{64}$
• B. $\frac{1}{16}$
• C. $64$
• D. $16$

Answer 1

Answer: (A), (C)

$64$

Given
${\log }_{16}x+{\log }_{x}16={\log }_{512}x+{\log }_{x}512\Rightarrow \frac{1}{4}{\log }_{2}x+4{\log }_{x}2=\frac{1}{9}{\log }_{2}x+9{\log }_{x}2$
Assuming ${\log }_{2}x=t$, we get
$\Rightarrow \frac{t}{4}+\frac{4}{t}=\frac{t}{9}+\frac{9}{t}\Rightarrow \frac{t}{4}-\frac{t}{9}=\frac{9}{t}-\frac{4}{t}\Rightarrow \frac{5t}{36}=\frac{5}{t}\Rightarrow t^2=36\Rightarrow t=\pm 6\Rightarrow {\log }_{2}x=\pm 6\Rightarrow x=\frac{1}{64},64$