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)

The correct options are
A 7
B 13

${\log }_{7}x+{\log }_{13}x=1$
$\frac{{\log }_{e}x}{{\log }_{e}7}+\frac{{\log }_{e}x}{{\log }_{e}13}=1$
$\log x[\frac{1}{\log 7}+\frac{1}{\log 13}]=1$
$\log x=\frac{\log 7.\log 13}{\log 7+\log 13}$
$\log x=\frac{\log 7.\log 13}{\log 91}⟶1$
$x={13}^{{\log }_{k}7}$
$\log x={\log }_{k}7\log 13$
$\frac{\log 7.\log 13}{\log 91}=\frac{\log 7}{\log k}\log 13$
$\log k=\log 91$
$e^{\log k}=e^{\log 91}$
$k=91$
k is divided by both 7 and 13