Solve $\left(\frac{n}{e}\right)^n=10$ without using estimation.

Question

$$\left(\frac{n}{e}\right)^n=10,$$ where $e$ is Euler's number.

I am unsure how to solve this problem. Do I take the $\log$ of $10$?

I have estimated it to be around $4.525$.

Answer 1

Let $f(x)=x^x e^{-x}$ so $f'/f=(x\ln x-x)'=\ln x$. The Newton-Raphson method gives an iteration of the form $x_{n+1}=x_n-\dfrac{1-10(e/x_n)^{x_n}}{\ln x_n}$. Start with $x_0=5$.

Answer 2

Take logarithms and divide by $e$ \begin{eqnarray*} \frac{n}{e} \ln( \frac{n}{e}) =\frac{\ln(10)}{e} \\ e^{\ln(n/e)} \ln( \frac{n}{e}) =\frac{\ln(10)}{e} \\ \ln(n/e)=W(\ln(10)/e) \\ n=e^{1+W(\ln(10)/e)} = 4.52 \cdots \end{eqnarray*} where $W$ is the Lambert function https://en.wikipedia.org/wiki/Lambert_W_function. It can be evaluated using Wolfie https://www.wolframalpha.com/input/?i=e%5E%7B1%2BProductLog(ln(10)%2Fe)%7D

Answer 3

The quickest way to solve this numerically is to rearrange the equation $$\left(\frac xe\right)^x=10$$ into the iterative sequence $$x_{n+1}=e\times10^{\frac{1}{x_n}}$$ and start with $$x_1=4$$

After a few steps you get $$x\simeq4.522724579...$$