Which is greater ${\log }_{7}10$ or ${\log }_{11}13$?

The correct option is A ${\log }_{7}10$
Let $y={\log }_{7}10-\frac{{\log }_{7}13}{{\log }_{7}11}=\frac{{\log }_{7}10.{\log }_{7}11-{\log }_{7}13}{{\log }_{7}11}$
$=\frac{{\log }_7\left(7.\frac{10}{7}\right){\log }_7\left(7.\frac{11}{7}\right)-{\log }_7\left(7.\frac{13}{7}\right)}{{\log }_{7}11}=\frac{(1+{\log }_7\frac{10}{7})(1+{\log }_7\frac{11}{7})-(1+{\log }_7\frac{13}{7})}{{\log }_{7}11}$
$=\frac{{\log }_7\frac{110}{49}+{\log }_7\frac{10}{7}{\log }_7\frac{11}{7}-\log \frac{13}{7}}{{\log }_{7}11}=\frac{{\log }_7\frac{110}{91}+{\log }_7\frac{10}{7}{\log }_7\frac{11}{7}}{{\log }_{7}11}>0$
Hence, ${\log }_{7}10$ is greater than ${\log }_{11}13$.