Lauren sent in her solution:

Since log_b x=log_b(y^{log_y x})=log_y xlog_b y, we have

logy x=

logb x logb y

and (letting b=x) 1=log_y x log_x y.
Now

log5 49=

log7 49 log7 5

=

2 log7 5

and

log7 125=

log5 125 log5 7

=

3 log5 7

, so

log5 49 ×log7 125=

2 log7 5

×

3 log5 7

=6

.