I have noticed that:
- The equation $x\cdot0.96^x=2.5$ has two real solutions
- The equation $x\cdot0.96^x=10$ has no real solutions
The full results on WolframAlpha are given below (or you can click each one of the links above).
My question is with regards to the general equation $xp^x=q$, with both $p$ and $q$ being positive:
How can I determine - as a function of $p$ and $q$ of course - if $xp^x=q$ has any real solutions?
From the research that I have done so far, I believe that the answer is:
$xp^x=q$ has real solutions $\iff p=1$ or $\frac{W(q\log(p))}{\log(p)}$ has real solutions, where $W$ is the Lambert W Function.
I'm not sure if I'm right or wrong here, so please let me know if I'm wrong.
If I'm right, then my question can be reduced to:
How can I determine - as a function of $p$ and $q$ of course - if $W(q\log(p))$ has any real solutions?
From what I read on Wikipedia (see link above), I have concluded that:
$W(q\log(p))$ has real solutions $\iff q\log(p)\geq-1/e$
Is that correct?
Thank you very much for reading through!
Hint
Consider that you look for the zero's of $$f(x)=xp^x-q$$ for which $$f'(x)=p^x (x \log (p)+1)\qquad \text{and}\qquad f''(x)=p^x \log (p) (x \log (p)+2)$$ The first derivative cancels at $$x_*=-\frac{1}{\log (p)}$$ for which $$f(x_*)=-\frac{1}{e \log (p)}-q\qquad \text{and}\qquad f''(x_*)=\frac{\log (p)}{e}$$
Now, analyze what happens if $p>1$ or $p<1$ considering that $x_*$ could be a minimum or a maximum.