If log(x-6) base 3=2y and log(x-7)base 2 =3y, show that x^2-13x+14=72^y

Answer 1

rewriting the logs, we get
3^(2y) = x-6
2^(3y) = x-7
so
9^y = x-6
8^y = x-7
and thus
72^y = (x-6)(x-7) = x^2-13x+42