$\ln(x)=ax^4$. Find a value for $a$ such that the function has only one real root.

How do i go about this kind of exercises? It is for a multiple choice exam for college admission so i only need the reason behind it in order to be able to solve such exercises. I tried plotting the graphs but I cant tell.


closed as off-topic by Namaste, Mostafa Ayaz, Cameron Williams, José Carlos Santos, mechanodroid Jul 14 '18 at 23:51

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  • $\begingroup$ Are we assuming $a > 0$. Other wise $a = 0$ will yield $\ln x = 0$ has only one solution. $\endgroup$ – fleablood Jul 14 '18 at 18:10

Look at the figure below

enter image description here

the curvature of $ax^4$ is always positive and that of $\ln x$ is always negative (by twice differentiating them) therefore they intersect in exactly one point if they are also tangent in that point i.e.$$\ln x=ax^4\\\dfrac{1}{x}=4ax^3$$which leads to $$ax^4={1\over 4}$$ and by substituting we have $$\ln x=\dfrac{1}{4}$$which yields to $$x=e^{1\over 4}$$and $$a=\dfrac{1}{4e}$$also the case $a\le 0$ is trivial since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $\ln x$ and $ax^4$ are strictly increasing and decreasing respectively.

  • $\begingroup$ What if $a \le 0$? $\endgroup$ – fleablood Jul 14 '18 at 18:11
  • $\begingroup$ This is the trivial case since for $a=0$ the only answer is $x=1$ which is valid and for $a<0$ the functions $\ln x$ and $ax^4$ are strictly increasing and decreasing respectively. $\endgroup$ – Mostafa Ayaz Jul 14 '18 at 18:15


Obviously, $x>0$. $\ln x=ax^4$ could be rewritten as $$a=\frac{\ln x}{x^4}.$$ We want to find the range of $a$ such that $y=a$ and $y=f(x)=\dfrac{\ln x}{x^4}$ intersect at only one point.

Notice that $$f'(x)=\frac{1-4\ln x}{x^5}.$$ Let $y'=0$. We have $x=e^{\frac{1}{4}}.$ Morover, $f'(x)>0$ when $x<e^{\frac{1}{4}}$ and $f'(x)<0$ when $x>e^{\frac{1}{4}}.$ Thus, we may claim that $f(x)$ increases over $(0,e^{\frac{1}{4}}]$ and decrease over $[e^{\frac{1}{4}},+\infty].$ Futher, $f(x) \in (-\infty,\dfrac{1}{4e}]$ and $f(x) \in (\dfrac{1}{4e},0)$. Here, you should notice $$\lim\limits_{x \to +\infty}\frac{\ln x}{x^4}=\lim\limits_{x \to +\infty}\frac{1/ x}{4x^3}=\lim\limits_{x \to +\infty}\frac{1}{4x^4}=0.$$

Now, we may obtain that the range of $a$ we want is $(-\infty,0]\bigcap\{\dfrac{1}{4e}\}$. Otherwise, there exist either two or none intersetion.

  • $\begingroup$ Thank you very much. I actually understand the reason behind such exercises but I could not figure how to start it. I got confused because a was not separated from x and could not understand why. Thank you very much, again, and everyone else who has commented. $\endgroup$ – Radu Gabriel Jul 14 '18 at 18:36
  • $\begingroup$ Not at all. please correct me if I'm wrong. $\endgroup$ – mengdie1982 Jul 14 '18 at 18:45

It is easy to show that the function $\ln x-ax^4$ has at most a single extremum, at $x=\dfrac1{\sqrt[4]{4a}}$.

For any $a\le0$, there is no extremum and a single root, as the function goes from $-\infty$ to $\infty$.

If $a>0$, the value at the extremum is

$$\frac14\ln4a-\frac14$$ and is zero for $4a=e$, forming a double root.


Assume $a >0$.

Considering the U/valley shape of the graph of $ax^4$ and the slouching slug shape of $\ln x$ it's clear that the graphs either never intersect and $\ln x < ax^4$ for all $x> 0$. Or they touch once and there is one solution to $\ln x = ax^4$ or the cross and recross and there are two solutions.

Now to solve $\ln x = ax^4$ is to solve $h(x) = ax^4 - \ln x = 0$ and as $a^4 \ge \ln x$ (the graphs touch once and never cross) then $h(x) = 0$ is local minimum.

So $h(x) = ax^4 - \ln x = 0$ and $h'(x) = 4ax^3 - \frac 1x =0$. Then later becomes $4ax^4 - 1 = 0$ and notice both $ax^4 - \ln x = 0$ and $4ax^4 - 1=0$ have terms pertaining to $ax^4$. And we can solve $ax^4 = \frac 14$.

So $\frac 14 - \ln x = 0$ and $\ln x = \frac 14$ and $x = e^{\frac 14}$.

Plugging that in we get $a*e - \frac 14 = 0$.

Which gives us $a = \frac 1{4e}$

If $a =0$ then $\ln x = ax^4 = 0$ and $x = 0$ is the onlys solution.

If $a < 0$. Well, look at the graphs. The U valley is upside down and every value intersect the slouching slug exactly once. So $a \in (-\infty, 0]\cup \{\frac 1{4e}\}$.


Hint: Consider the function $f(x)=\frac{\ln(x)}{x^4}$ and use calculus. Use that $$f'(x)=\frac{1-4\ln(x)}{x^4}$$

  • 1
    $\begingroup$ don't you mean $f'(x)=\frac{1-4ln(x)}{x^5}$ $\endgroup$ – Mandelbrot Jul 14 '18 at 17:17

\begin{align} \ln x&=ax^4 \tag{1}\label{1} ,\\ -4\ln x&=-4ax^4 ,\\ \ln(x^{-4})&=-4ax^4 ,\\ x^{-4}\ln(x^{-4})&=-4a . \end{align}

Let $\ln(x^{-4})=u$, then $x^{-4}=\exp(u)$ and we arrive at an equation

\begin{align} u\exp(u)&=-4a , \end{align}

which has a well-known closed-form solution in terms of Lambert W function.

\begin{align} \operatorname{W}(u\exp(u))&=\operatorname{W}(-4a) ,\\ u&=\operatorname{W}(-4a) ,\\ \ln(x^{-4})&=\operatorname{W}(-4a) ,\\ x^{-4}&=\exp(\operatorname{W}(-4a)) ,\\ x&=\exp(-\tfrac14\cdot\operatorname{W}(-4a)) \tag{2}\label{2} . \end{align}

The number of real solutions to \eqref{1} is defined by the number of real solutions to \eqref{2}, which in turn, is defined by the argument of $\operatorname{W}$, and it is well-known that we have a single real solution when the argument of $\operatorname{W}$ is either non-negative, or is equal to $-\tfrac1{\mathrm{e}}$.

That is,

\begin{align} 1)& -4a\ge0\quad\to\quad a\le0 ;\\ 2)& -4a=-\tfrac1{\mathrm{e}} \quad\to\quad a=\tfrac1{\mathrm{4e}} \approx 0.09196986 . \end{align}

In the last case we have

\begin{align} x&=\exp(-\tfrac14\cdot\operatorname{W}(-\tfrac1{\mathrm{e}})) =\exp(-\tfrac14\cdot(-1)) =\sqrt[4]{\mathrm{e}} . \end{align}


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