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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.

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  • $\begingroup$ 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. $\endgroup$
    – Community Bot
    Feb 3 at 11:53
  • $\begingroup$ If you plot $e^x-x^4,e^x$ dominates after about $x=8.613$ $\endgroup$ Feb 3 at 12:25
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    $\begingroup$ You can change $e^x=x^n$ into $x/lnx=n$ which is a more distinct way. $\endgroup$ Feb 3 at 12:52

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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 $x^n$.

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  • $\begingroup$ Actually I can use calculus but this ques came in our assignment which was of functions also teacher restricted us from using calculus . $\endgroup$
    – Macron
    Feb 3 at 12:58
  • $\begingroup$ @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 $e^{10} > 10^4$. $\endgroup$ Feb 3 at 13:05
  • $\begingroup$ yes i got that thank you $\endgroup$
    – Macron
    Feb 3 at 13:21
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    $\begingroup$ @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? $\endgroup$
    – David K
    Feb 3 at 14:50
  • $\begingroup$ @DavidK in high school all real numbers are defined by their decimal representation. It's why we get so many silly questions about it. $\endgroup$ Feb 3 at 18:18
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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.

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