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.

  • $\begingroup$ In general, you cannot find exact intersection points of a polynomial and an exponential function. You might want to read up on Newton’a approximation method and Bisection Method $\endgroup$ – Sina Babaei Zadeh Dec 9 '19 at 2:54
  • $\begingroup$ So there is no way I can determine the number of intersection points of x^4 and 4^x using high school calculus. $\endgroup$ – Abhinav Gupta Dec 9 '19 at 2:56
  • $\begingroup$ You should be well equipped enough to at least attempt newton's approximation with high school calculus, but there are no algebraic methods for you to "get x by itself" and solve for the negative intersection points. x is just not separable in this case. $\endgroup$ – Ty Jensen Dec 9 '19 at 3:02
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    $\begingroup$ You can’t find the exact solution using high school calculus but most people consider the Bisection Method to be at a precalculus level and it gives a decent approximation. With this method, you write your function as $x^4-4^x=y$ and find intervals where your function changes signs. Then you start to make than interval smaller until you get a decent approximation. For example, between $3$ and $1$ your function $x^4-4^x=y$ changes sign which implies that there is a point of intersection between those values and you can make this interval smaller by taking midpoints. Many approximations exist. $\endgroup$ – Sina Babaei Zadeh Dec 9 '19 at 3:09

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.


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.

  • $\begingroup$ In "This has exactly one extreme value", This refers to $f$, not $f'$. $\endgroup$ – Gerry Myerson Dec 9 '19 at 3:30
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    $\begingroup$ @GerryMyerson Thank you for pointing that out. I fixed my post. $\endgroup$ – Matthew Daly Dec 9 '19 at 3:48

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