Log Products

$(\log_2{3})(\log_3{4})(\log_4{5})(\log_5{6})...(\log_{511}{512}) \\ = \left( \dfrac{\log{3}}{\log{2}} \right) \left( \dfrac{\log{4}}{\log{3}} \right) \left( \dfrac{\log{5}}{\log{4}} \right) \left( \dfrac{\log{6}}{\log{5}} \right) ... \left( \dfrac{\log{512}}{\log{511}} \right) \\ = \left( \dfrac{\log{512}}{\log{2}} \right) = \left( \dfrac{\log{2^9}}{\log{2}} \right) = \left( \dfrac{9\log{2}}{\log{2}} \right) = \boxed{9}$

Let the given expression equal $x.$ Then since $\large a^{\log_{a}(b)} = b$ we have that

$\large 2^{x} = (2^{\log_{2}(3)})^{(\log_{3}(4))(\log_{4}(5))....(\log_{511}(512))} =$

$\large (3^{\log_{3}(4)})^{(\log_{4}(5))(\log_{5}(6))....(\log_{511}(512))} =$

$\large (4^{\log_{4}(5)})^{(\log_{5}(6))(\log_{6}(7))....(\log_{511}(512))}$

and so on, creating a "domino effect" which leaves us with

$\large 2^{x} = 511^{\log_{511}(512)} = 512 \Longrightarrow x = \boxed{9}.$

The property used here is log b to the base a can be written as (log b)/(log a).

So, log 3 to the base 2 is (log 3)/(log 2).

All the terms cancel in this telescopic multiplication and the remaining terms are log(512)/log(2),

Also, log(a^b) = b x log(a).

512 = 2^9.

So, 9 log(2)/log(2) = 9.