A Series of Logarithms

Rewrite using change of base property: since $\log _{ a }{ b } =\frac { \log { b } }{ \log { a } }$ , we can conclude that $\frac { 1 }{ \log _{ a }{ b } } =\log _{ b }{ a }$

Thus we can rewrite the entire expression as $\log _{ x }{ 2 } +\log _{ x }{ 3 } ...\log _{ x }{ 2015 }$ .

By the sum of logs property, $\log _{ x }{ 2 } +\log _{ x }{ 3 } ...\log _{ x }{ 2015 } =\log _{ x }{ (2\times 3\times 4...2015)=\log _{ x }{ (2015!) } }$

Because $x = 2015!$ , $\log_{x}{(2015!)} = 1$