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Solve:

$$x^x = \ln x + 1$$

There should be only one solution, $ x=1$, but I cannot prove it.

lulu
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Michael Lawrence
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    You can easily verify the $x = 1$ solution by just plugging it in. Proving that there aren't any other solutions is a bit harder. – Dan Nov 14 '22 at 19:48
  • The derivative is monotone increasing and $x=1$ is a global min. – lulu Nov 14 '22 at 19:57
  • Can you want to improve your question? For example, what have you tried? – lone student Nov 14 '22 at 20:01
  • Use the well-known inequality: $e^x≥x+1$.

    You have:

    $$ \begin{align}\ln x+1=x^x=e^{x\ln x}&≥x\ln x+1\ \ln x+1&≥x\ln x+1\ \ln x &≥x\ln x\ \ln x(x-1)&≤0. \end{align} $$

    Now, you can complete the final step.

     – lone student Nov 14 '22 at 20:10
  • $f(x)=x^x$ is a convex function on $]0,+\infty[$ and $g(x)=\ln x+1$ is a concave function on $]0,+\infty[$. Since the curves $y=f(x)$ and $y=g(x)$ have the same tangent at the point $P(1,1)$ that is $y=x$, it follows that $P(1,1)$ is the only point of intersection of them, hence $x=1$ is the only solution of the equation $f(x)=g(x)$. – Angelo Nov 14 '22 at 21:18

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