Finding parameter so that a equation has $ 3$ solution.

Question

I am trying to find a real number $a$ so that $$|\log x|= ax$$ has $3$ distinct solutions. Playing with the graph I found that a must be in $(0, e^{-1})$, and now I am trying to show analytically. If we note $h(x)=\frac{|\log x|}{x}$ then $$h'(x)=\frac{\log x- \log^2 x }{x^2|\log x|}$$ So there is only a global minimum at $x=1$, now I don't see how I can show it have 3 solutions and I think I'm on the wrong track. Could you help me solve this problem?

Answer 1

Calculate the tangent on $\log x$ from point $(0,0)$. Say that tangent has a slope $k$. Thus $a\in (0,k)$

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Say that tangent touch $f(x) = \log x$ at $T(b,c)$. Then $k=f'(b) = 1/b$ and $c=f(b)=\log b$. So the equation of tangent at $T$ is $$y-c={1\over b}(x-b)$$

But since it goes through $(0,0)$ we have $-c =-1$, so $c=1$ and so $b=e$ so $k=1/e$ and so $\boxed{a\in (0,{1\over e})}$.

Answer 2

If $0 < a < 1/e$, then $\log(x) = -ax$ for $x = W_0(a)/a$ and $\log(x) = ax$ for $x = -W_0(-a)/a$ and $x = -W_{-1}(-a)/a$, where $W_0$ and $W_{-1}$ are branches of the Lambert W function.