How many points in the xy-plane do the graphs of $y=x^{12}$ and $y=2^x$ intersect? The question in the title is equivalent to find the number of the zeros of the function $$f(x)=x^{12}-2^x$$
Geometrically, it is not hard to determine that there is one intersect in the second quadrant. And when $x>0$, $x^{12}=2^x$ is equivalent to $\log x=\frac{\log 2}{12}x$. There are two intersects since $\frac{\log 2}{12}<e$. 

Is there other quicker way to show
  this?


Edit: This question is motivated from a GRE math subject test problem which is a multiple-choice one(A. None B. One C. Two D.Three E. Four). Usually, the ability for a student to solve such problem as quickly as possible may be valuable at least for this kind of test. In this particular case, geometric intuition may be misleading if one simply sketch the curve of two functions to find the possible intersect.     

This is not really an independent answer, more a gathering in one place of what others have said. 
It's easy to see there's a solution between $-1$ and $0$ (since $x^{12}-2^x$ is positive at $-1$ and negative at $0$), and another solution between $1$ and $2$ (again, there's a change of sign), and another solution bigger than $2$ (by Theo Buehler's comment on Joseph Malkevitch's answer). Then there are any number of ways to show that there are no solutions other than these three. 
