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What is the approximate solution to:

$x ^{1000} = 1000 !$

How can you solve for x?

More generally, for some constant $k$, how can you solve:

$x^k = k!$

Asaf Karagila
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Andrew Tomazos
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  • As $1000!$ has many, many prime factors, it is impossible to get the exact number without getting too large. But there might be approximate solutions. – Mathemagician314 Feb 07 '21 at 12:37
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    When we approximate $\sum_{k=1}^n\ln k$ with $\int_1^k\ln x,dx$ we get the approximate formula $\ln (k!)\approx k\ln k-k$. Therefore $$\ln x=\ln \root k\of{k!}\approx\frac1k(k\ln k-k)=\ln k-1.$$ So $x\approx k/e$. With $k=1000$ we have $1000/e\approx 367.88$. The exact solution would be $x\approx369.49$. Expect approximation error in that ballpark. – Jyrki Lahtonen Feb 07 '21 at 12:39
  • The upper limit of my intergal should be $n$ rather than $k$. Or, possible better, the summation should be $\sum_{i=1}^k\ln i$. Too late to edit. Sorry. – Jyrki Lahtonen Feb 07 '21 at 12:45
  • First of all , $\sqrt[k]{k!}$ is the exact solution. For large $k$ , we can use Stirling as mentioned but we can also just multiply the $k$-th roots of the numbers from $1$ to $k$. The solution is irrational for every $k\ge 2$, so rounding is inevitable anyway, if we want to determine the decimal expansion. – Peter Feb 07 '21 at 14:00
  • If $k$ is too large to calculate the product , or if you do not have a table calculator that can handle loops , or the computation would be too time consuming, then we must be content with the Stirling approximation. – Peter Feb 07 '21 at 14:01

2 Answers2

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You can use stirling's approximation when $n \rightarrow +\infty$: $$\ln(n!) = n\ln(n)-n+O(\ln(n))$$

So, \begin{align} x^k = k! & \Leftrightarrow k\ln(x) =\ln(k!) \\ & \Leftrightarrow k\ln(x) =k\ln(k)-k+O(\ln(k)) \\ & \Leftrightarrow \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\ & \Leftrightarrow \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\ & \Leftrightarrow x =\frac{k}{e} \times\exp{(O(\frac{\ln(k)}{k}))} \approx \frac{k}{e}\\ \end{align}

NN2
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Notice that by the Hierarchy of limits, $\frac{x^n}{n!}\rightarrow 0$ as $n\rightarrow\infty$, so $n$ would have to be 'quite small'. In response to your first problem, $1000\log{x}=log{1000!}$ so $\log{x}=\frac{log{1000!}}{1000}$. Can you see a way to complete the problem? If not, here is a hint:

What can you deduce about the function $f(x)=\frac{log{x!}}{x}$? Can you sketch a graph of it? How does its derivative compare with $\log{x}$? You should be able to use these to find an approximate intersection point.

[To differentiate ${\log{x!}}$, consider the Gamma function definition of factorials, then apply the Fundamental Theorem of Calculus to differentiate.]

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    How would you differentiate your function $f$? What does $x!$ mean for noninteger $x$? – Randall Feb 07 '21 at 12:45
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    Yes. That might be worth including. – Randall Feb 07 '21 at 12:46
  • We can use the gamma function for factorials I believe - then use Fundamental Theorem of Calculus to differentiate –  Feb 07 '21 at 12:47
  • Thanks for pointing this out Randall - I'll include this now:) –  Feb 07 '21 at 12:47
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    I ask because differentiating the gamma function is highly nontrivial. – Randall Feb 07 '21 at 13:37
  • I agree it's not exactly standard but the formula for the $n$th derivative is relatively straightforward so could be memorised. –  Feb 07 '21 at 13:44
  • I don’t know any nice formulation of its derivative: https://math.stackexchange.com/q/1634781/464495 – Randall Feb 07 '21 at 13:51