$\ln x = ax^4$

I found a solution using graph. Another method exists, eventually using analysis?

Omnomnomnom 07/11/2018.

We can rearrange the equation to say that we're looking for an $a$ such that $$ \frac{\ln(x)}{x^4} = a $$ has exactly one solution. In other words: if we look at the graph of $y = f(x) = \ln(x)/x^4$, we are looking for the $y$-values that are attained exactly once.

In order to get the important features the graph of $f(x)$, it suffices to observe that:

- $\lim_{x \to 0^+} f(x) = -\infty$
- $\lim_{x \to \infty} f(x) = 0$
- $f(x)$ is increasing on $(0,e^{1/4})$ and decreasing on $(e^{1/4},\infty)$.

These facts can be verified by the usual "intro calculus" methods.

Since $f$ attains a maximum at $x = e^{1/4}$, we see that $f(e^{1/4}) = \frac 1{4e}$ is the only positive value of $a$ that is attained exactly once.

fleablood 07/11/2018.

Take $f(x) = ax^4 - \ln x$. We want that to equal $0$ exactly once.

Informally in terms of graphs that means $f(x)$ "touches" $y=0$ once but does not "cross" it. So $f(x) = 0$ is a local maximum or minimum.

So we need an $a$ and an $x$ so that $f(x) = 0$ and $f'(x) = 0$.

So $f'(x) = 4ax^3 - \frac 1x = 0$

$4ax^4 - 1 = 0$

$ax^4 = \frac 14$

$x = \pm (\frac 1{4a})^{\frac 14}$ So $a > 0$.

We also have $ax^4 - \ln x = 0$

So $\ln x = ax^4 = \frac 14$

So $x = e^{\frac 14}$ and $x = \pm (\frac 1{4a})^{\frac 14}$ (we can note that this means that $x$ is positive but that's not really that important as it is the $a$ we are interested in.)

So $x^4 = e = \frac 1{4a}$ and $a = \frac 1{4e}$

J.G. 07/11/2018.

Set $ax^4-\ln x$ and its derivative equal to $0$, so the root is a turning point. You get $4ax^3=1/x$ and $\ln x=ax^4=\frac{1}{4}$, so $a=\frac{1}{4}(\exp\frac{1}{4})^{-4}=\frac{1}{4e}$.

Robert Israel 07/11/2018.

As $x \to 0+$, $\ln(x) \to -\infty$ while $a x^4 \to 0$ so $\ln(x) < a x^4$ for $x$ sufficiently small. On the other hand, $ \ln(x) < a x^4$ for $x$ sufficiently large if $a > 0$. To have a single solution, you'll want the two graphs tangent. Since $\ln(x) - a x^4$ is strictly concave, this implies there is only one solution.

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