# Bounds for solutions of $xe^{x} -n =0$ for $n\geq 3$

$\alpha _{ n }$ is a solution for $f_{ n }=xe^{ x }-n =0$. How can I prove that $\forall n\geq 3$, $\ln(n)-\ln(\ln(n))\le \alpha _{ n }\le \ln(n)$

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## 3 Answers

For $x\ge-1$, $\frac{\mathrm{d}}{\mathrm{d}x}xe^x=(1+x)e^x\ge0$. Thus, $f(x)=xe^x$ is monotonic increasing for $x\ge-1$. Therefore, for $x\ge e$, $$\begin{align} f(\log(x)) &=\log(x)e^{\log(x)}\\ &=x\log(x)\\ &\ge x \end{align}$$ Furthermore, $$\begin{align} f(\log(x)-\log(\log(x))) &=(\log(x)-\log(\log(x)))e^{\log(x)-\log(\log(x))}\\ &=\frac{\log(x)-\log(\log(x))}{\log(x)}x\\ &=\left(1-\frac{\log(\log(x))}{\log(x)}\right)x\\ &\le x \end{align}$$ Therefore, for $x\ge e$, $$\log(x)-\log(\log(x))\le f^{-1}(x)\le\log(x)$$

Lambert W

$f^{-1}$ is commonly known as the Lambert W function.

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Note that $f_n(x)=0$ iff $g(x)=xe^x=n$.

$g(x)$ is monotone increasing, so $(g(x)>n) \Leftrightarrow (x>\alpha_n)$ and $(g(x)<n) \Leftrightarrow (x<\alpha_n)$

The upper bound: $g(\ln n) = n\ln n > n$, so as written above, $\alpha_n < \ln$.

The lower bound:

$$g(\ln n - \ln(\ln n)) = (\ln n - \ln(\ln n))e^{\ln n - \ln(\ln n)}$$ $$= (\ln n - \ln(\ln n))n\cdot \frac{1}{\ln n}$$ $$= \left(1 - \frac{\ln(\ln n)}{\ln n}\right)n < n$$

So again because $g(x)$ is monotone increasing, $\alpha_n>\ln n - \ln(\ln n)$.

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Let $\alpha_n$ be a solution for some $n\geq 3$. Then $$\alpha_n e^{\alpha_n}=n \iff \alpha_n+\log(\alpha_n)=\log(n),$$ and so if $\log(\alpha_n)>0$ or equivalently $\alpha_n>1$ (which is true since $n\geq 3$), then $\alpha_n\leq \log(n)$ by the above.

Now, since $\alpha_n\leq \log(n)$ and $x\mapsto \log(x)$ is increasing, then $\log(\alpha_n)\leq\log(\log(n))$ and so we obtain the lower bound: $$\alpha_n=\log(n)-\log(\alpha_n)\geq \log(n)-\log(\log(n)).$$

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