Is there a way to know how many times two curves intersect? I want to know how many times two curves intersect. I know we can just solve them but the curves I am dealing with are $y=x^4$ and $y=5^x$. I don't know how to solve them together
I want to know if by using calculus we can plot a graph of the two of if we can see how many times they intersect. 
They will intersect once for sure for some negative value of $x$ but in the first quadrant of the Cartesian plane I am not sure if they will intersect or not.
 A: Consider $f(x)=5^xx^{-4}$.  You will have a solution to $x^4=5^x$ whenever $f(x)=1$, so we can do what we can to investigate how many solutions there are to that.  As you note, it is pretty clear that there is one solution in the second quadrant.  As for the first quadrant, let's calculate
$$f'(x)=5^xx^{-5}(x\ln5-4)$$
This has exactly one zero, which corresponds to a local minimum of $f$ when $x=\frac4{\ln5}$.  In that case, $f(x)\approx1.431$.  Therefore, there is no solution to $x^4=5^x$ when $x$ is positive.
A: You can find the number of intersections without finding out the precise locations of the intersections. Let $f(x)=x^4$, $g(x)=5^x$. $f(-1)>g(-1)$, $f(0)<g(0)$ so there's a solution between $-1$ and zero. Now we want to show that's the only solution. We take derivatives. $f'(x)=4x^3$, $g'(x)=5^x\log5$. We have $f(0)<g(0)$, so if we can prove $f'(x)<g'(x)$ for all $x>0$, then we can conclude $f(x)<g(x)$ for all $x>0$. Has this made the problem any easier? Well, we've replaced $f$ with $f'$, and $f'<f$ for $x>4$. If we repeat the procedure a few more times, we'll get to where we need to show $0<5^x(\log 5)^5$, which is trivial. 
You need to be a little careful around the edges of this sketch, but you should have a go at filling in the details. 
