# Solving a particularly difficult exponential equation

I was recently looking for the maximum value of $x^{10-x}$ over $x>0$ without using calculus. I first thought of expanding it to $\frac{x^{10}}{x^x}$ and then looking for when $x^x$ grows faster or slower than $x^{10}$ by incrementing $x$ by 1 (I know, not very rigorous). From this, I could deduce over which intervals the function was increasing or decreasing, and from this, the maximum value. After some math, I arrived at the inequalities $$x^{10-x}>{(x+1)}^{9-x}$$ and $$x^{10-x}<{(x+1)}^{9-x}$$ This is where I got stuck. Is there any way to solve these two inequalities algebraically? The only option I could think of is finding an approximation through a graph, but of course an exact answer would always be more useful. Thanks!

## 1 Answer \begin{align} f(x)=x^{10-x} ,\\ f'(x)=x^{9-x}(10-x-x\ln x) . \end{align}

Since $x^{9-x}>0$, $f'(x)=0$ when $10-x-x\ln x=0$, and this equation has one real solution in terms of Lambert W function as follows:

\begin{align} x+x\ln x&=10 ,\\ x(1+\ln x)&=10 ,\\ x(\ln \mathrm{e}\,x)&=10 ,\\ \mathrm{e}\,x(\ln \mathrm{e}\,x)&=10\,\mathrm{e} ,\\ (\ln \mathrm{e}\,x)\,\exp(\ln \mathrm{e}\,x)&=10\,\mathrm{e} ,\\ \ln \mathrm{e}\,x&=\operatorname{W}(10\,\mathrm{e}) ,\\ x&=\exp(\operatorname{W}(10\,\mathrm{e})-1) \approx 4.13366 ,\\ f_{\max}=f(\exp(\operatorname{W}(10\,\mathrm{e})-1)) &\approx 4126.9500 . \end{align}