Question

If ${\log }_{7}x+{\log }_{13}x=1$ and $x={13}^{{\log }_{k}7}$, then $k$ is divisible by

• A. $7$
• B. $13$
• C. $17$
• D. $119$
  

Answer 1

Answer: (A), (B)

$13$

Given : ${\log }_{7}x+{\log }_{13}x=1$ and $x={13}^{{\log }_{k}7}$
Now,
$x={13}^{{\log }_{k}7}\Rightarrow {\log }_{13}x={\log }_{k}7x=7^{{\log }_{k}13}\Rightarrow {\log }_{7}x={\log }_{k}13$
Now,
${\log }_{7}x+{\log }_{13}x=1\Rightarrow {\log }_{k}7+{\log }_{k}13=1\Rightarrow {\log }_{k}91=1\therefore k=91$