Solving $x^{\left\lfloor x \right\rfloor}\; =\; 2014$ 
$x^{\left\lfloor x \right\rfloor}\; =\; 2014$

Mathematica gives that there are no solutions, but how do you actually come to the conclusion that there exists no solution to this equation?
 A: Suppose $x < 0$.  Then the LHS will have a strictly negative integer exponent:  $\lfloor x \rfloor = k$, where $k \in \{ -1, -2, \ldots \}.$  Hence we would require $x = (2014)^{1/k}$, but this number is positive, so we arrive at a contradiction. Thus $x > 0$. 
Indeed, we must have $x > 1$, for if $0 < x \le 1$, then the LHS is obviously $1$.  Now observe that because $x-1 < \lfloor x \rfloor \le x$, we must have $$x^{x-1} < x^{\lfloor x \rfloor} \le x^x.$$  Hence by direct calculation, we find $$4 < x < 5.$$  That is, $x = 4 + \epsilon$ for $\epsilon \in (0,1)$.  Then we have $$x^{\lfloor x \rfloor} = (4+\epsilon)^4.$$  If we require this to equal $2014$, then solving for $\epsilon$ immediately yields $$\epsilon = (2014)^{1/4} - 4 \approx 2.69908 \not\in (0,1),$$ hence no such solution exists.
Without calculating the numerical value of $\epsilon$, we can also easily determine that it exceeds $1$, for $2014 > 1296 = 6^4$, hence $(2014)^{1/4} > 6$.
A: Well, one way is to plot it, and you will see that the curve is discontinuous and jumps over 2014.
                                    


Of course this function is always increasing, so the discontinuity here is enough to show that there is no solution :)
