For every real number $x$, let $\lfloor x\rfloor$ denote the greatest integer not exceeding $x$, and let \[f(x)=\lfloor x\rfloor(2014^{x-\lfloor x\rfloor}-1).\]  The set of all numbers $x$ such that $1\leq x<2014$ and $f(x)\leq 1$ is a union of disjoint intervals.  What is the sum of the lengths of those intervals?
$\textbf{(A) }1\qquad \textbf{(B) }\dfrac{\log 2015}{\log 2014}\qquad \textbf{(C) }\dfrac{\log 2014}{\log 2013}\qquad \textbf{(D) }\dfrac{2014}{2013}\qquad \textbf{(E) }2014^{\frac1{2014}}\qquad$
Let $\lfloor x\rfloor=k$ for some integer $1\leq k\leq 2013$.  Then we can rewrite $f(x)$ as $k(2014^{x-k}-1)$.  In order for this to be less than or equal to $1$, we need $2014^{x-k}-1\leq\dfrac1k\implies x\leq k+\log_{2014}\left(\dfrac{k+1}k\right)$.  Combining this with the fact that $\lfloor x\rfloor =k$ gives that $x\in\left[k,k+\log_{2014}\left(\dfrac{k+1}k\right)\right]$, and so the length of the interval is $\log_{2014}\left(\dfrac{k+1}k\right)$.  We want the sum of all possible intervals such that the inequality holds true; since all of these intervals must be disjoint, we can sum from $k=1$ to $k=2013$ to get that the desired sum is \[\sum_{i=1}^{2013}\log_{2014}\left(\dfrac{i+1}i\right)=\log_{2014}\left(\prod_{i=1}^{2013}\dfrac{i+1}i\right)=\log_{2014}\left(\dfrac{2014}1\right)=\boxed{1\textbf{ (A)}}.\]

Video Solution by Richard Rusczyk
[https://artofproblemsolving.com/videos/amc/2014amc12a/380](https://artofproblemsolving.comhttps://artofproblemsolving.com/videos/amc/2014amc12a/380)
~ dolphin7