# Finding number of solutions of $e^x$ = $x^3$

As we can see $$e^x=x^2$$ has 1 solution.

$$e^x=x^4$$ has 2 solutions with $$x^4 > e^x$$ after $$x=1.43$$ which is one of the solutions.

What is the reason behind this?

Also what would be the approach is we were asked to find the number of solutions of $$e^x=x^3$$? If we see graph of $$e^x = x^3$$ we think it has 1 solution but it actually has 2 solutions $$x=1.857$$ and $$x=4.536$$, so why is there such variation in slopes?

As $$x^4$$ dominates over $$e^x$$ after some $$x$$ till infinity, while in $$x^2$$ and $$x^3$$, $$e^x$$ is dominating.

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Feb 3 at 11:53
• If you plot $e^x-x^4,e^x$ dominates after about $x=8.613$ Feb 3 at 12:25
• You can change $e^x=x^n$ into $x/lnx=n$ which is a more distinct way. Feb 3 at 12:52

## 2 Answers

Non-constant polynomials will only dominate $$y^x$$ when $$y \leq 1$$ and since $$e>1$$ we know that $$e^x$$ will dominate $$x^n$$ eventually. Note that your axis only goes as high as $$x=6$$ which is extremely far from $$\infty$$ so you can't tell anything about the dominating behavior. In fact, no graph of finite size will ever give you any information about the limit at $$\infty$$ and you'll need to rely on calculations to evaluate it.

If you graph $$e^n-x^n$$ it's zeroes will show you when $$e^x$$ begins to dominate $$x^n$$ as it crosses the $$x$$-axis. Here you can just check that $$e^{10} > 10^4$$ directly or you can just zoom out on the graph so that the values at $$x=10$$ visible.

If you're familiar with calculus then you can see taking derivatives of $$e^x$$ will always give you $$e^x$$ but I can differentiate a degree $$n$$ polynomial $$n$$ times and will be constant. So this means the growth of $$e^x$$ will always eventually be faster than a polynomial and thus will eventually overtake $$x^n$$.

• Actually I can use calculus but this ques came in our assignment which was of functions also teacher restricted us from using calculus . Feb 3 at 12:58
• @Macron calculus is optional and you always find a rough estimate for the point when it overtakes by guess-and-test. That's how I found that $e^{10} > 10^4$. Feb 3 at 13:05
• yes i got that thank you Feb 3 at 13:21
• @Macron If you are forbidden from "using calculus" for this exercise, how can you say what the function $e^x$ is? All of the definitions of $e^x$ that I know involve a limit in some way or another. What definition of $e^x$ do you think your teacher is assuming you should use? Feb 3 at 14:50
• @DavidK in high school all real numbers are defined by their decimal representation. It's why we get so many silly questions about it. Feb 3 at 18:18

To decide of the number of zeros of $$\frac{x}{3}-\log x$$, observe that it is 0, 1 or 2, because of the concavity of $$x\mapsto \log x.$$ Since $$y=x/e$$ is tangent to the curve of the log, and since $$1/3<1/e,$$ the answer is 2.