If the product \(\frac{3}{2}*\frac{4}{3}*\frac{5}{4}*\frac{6}{5}*...*\frac{a}{b}=9\), what is the sum of \(a\) and \(b\)?
By scanning the numerator and the denominator, we can see common numbers. So, 3 gets cancels out, 4, 5, 6, up to b, an unknown number. After all these cancelations, we are left with \(\frac{a}{2}=9\). This means that a, is \(9*2=18\). To find b, we look at our previous fractions, noting that the denominator is one less than the numerator, so that leaves us with: \(b=17\).
Thus, the answer is \(18+17=\boxed{35}\).