jmc

The number $A{B}_7$ is $7A+B$, and the number $B{A}_5$ is $5B+A$, so $7A+B=5B+A$. Then $6A=4B$, so $3A=2B$. Then $B$ must be a multiple of 3. But $B$ is also a digit in base 5, so $B=3$, and $A=2$. The number is $7A+2=\boxed{17}$.