For what positive value of $c$ does the equation $\log(x)=cx^4$ have exactly one real root?

I think I should find a way to apply IMV and Rolle's theorem to $f(x) = \log(x) - cx^4$. I think I should first find a range of values for $c$ such that the given equation has a solution and then try to find one that gives only a unique solution. I have thought over it but nothing comes to my mind. Perhaps I'm over-complicating it.

Any ideas on how to proceed are appreciated.


If $f(x)=\ln(x)$ and $g(x) = cx^4$ has exactly $1$ solution, then we have a condition of tangency at their common point.

So $$\displaystyle \left[f'(x)\right]_{(x_{1},y_{1})} = \frac{1}{x_{1}}$$ and $$\displaystyle \left[g'(x)\right]_{(x_{1},y_{1})} = 4cx^3_{1}$$

Note at $P(x_{1},y_{1})$ these slopes are equal

So $$\displaystyle \left[f'(x)\right]_{(x_{1},y_{1})} = \left[g'(x)\right]_{(x_{1},y_{1})}\Rightarrow \frac{1}{x_{1}} = 4cx^3_{1}\Rightarrow x^4_{1} = \frac{1}{4c}$$

Now point $P(x_{1},y_{1})$ also lies on $f(x)$ and $g(x)$

So $$\displaystyle \ln(x_{1}) = cx^4_{1} = c\cdot \frac{1}{4c} = \frac{1}{4}$$

So we get $$\displaystyle \ln(x_{1}) = \frac{1}{4}\Rightarrow x_{1} = e^{\frac{1}{4}}$$

So put $\displaystyle x = \frac{1}{4}$ into $\ln(x_{1}) = cx^4_{1}\;,$ and we get $\displaystyle \frac{1}{4} = c\cdot e \displaystyle \Rightarrow c = \frac{1}{4e}$

Note that for all $c\le0\;,$ these two curves intersect each other at exactly one point, but we're only interested in values of $c$ for which $c>0$.

So our final solution is $\displaystyle c = \left\{\frac{1}{4e}\right\}$

  • $\begingroup$ $\ln(x) = 0\Rightarrow x = e^{0} = 1$ and $\displaystyle c= \frac{1}{2e}$ $\endgroup$ – juantheron Sep 8 '15 at 7:39
  • $\begingroup$ Put $x=1$ in $\ln(x)$ $\endgroup$ – juantheron Sep 8 '15 at 7:41
  • $\begingroup$ Oops, right. Fortunately the problem has excluded non-positive solutions for $c$. XD $\endgroup$ – user246836 Sep 8 '15 at 7:43
  • $\begingroup$ Sorry HZ I have edited by solution, You are Right, Ans $\displaystyle = \frac{1}{4e}$ $\endgroup$ – juantheron Sep 8 '15 at 7:45
  • $\begingroup$ Why must the slopes be equal at the intersection point? I ask since it is not guaranteed that two intersecting curves will have a common tangent line. Intersecting lines is an easy counter example. So why can we assume that the two curves in this problem will share a tangent line? $\endgroup$ – rosterherik Oct 26 '18 at 22:10

Hint: Let $x = k$ be the unique place at which both curves intersect. Then we also know that both curves have a common tangent line at this point (try sketching a few example curves to see why), so we can equate derivatives. Thus, we must solve the following system of equations (assuming $\log$ refers to the natural log $\ln$): \begin{cases} \ln k = ck^4 \\ \dfrac{1}{k} = 4ck^3 \end{cases}


There will be precisely one real root if and only if $c\leq 0$ or $c={1\over 4e}$. The details are below.

If $c\leq 0$, then the derivative ${1\over x}-4cx^3$ is strictly positive in $(x,\infty)$, and since the function $\log (x)-cx^4$ tends to $-\infty$ as $x\to 0$ (from the right) and tends to $+\infty$ as $x\to\infty$, you get that there is one real root, by continuity, and it is unique, because the function is strictly increasing.

If $c > 0$, the the derivative ${1\over x}-4cx^3$ has precisely one real root, (check this), and the function tends to $-\infty$ at both extremes, so the only way it will have a unique root is when the $x$ axis is tangent to the graph, i.e. when both the function and it's derivative are zero. Solving the equations gives $c={1\over 4e}$.

Added in proof: I did not notice that you originally asked for only the positive cases of $c$. This leaves you only with $c={1\over 4e}$.


Solution in terms of Lambert W function:

\begin{align} \log(x)&=cx^4, \\ x^{-4}\log(x^{-4})&=-4c, \\ \log(x^{-4})&=\mathop{W}(-4c), \\ x&=\exp\left(-\tfrac14\mathop{W}(-4c)\right). \end{align}

For $c>0$ there is only one point $t=-\exp(-1)$, where the two real branches $\mathop{W_0}(t)$ and $\mathop{W_{-1}(t)}$ met, so the answer is $c=\tfrac14\exp(-1)$.

  • $\begingroup$ How did you arrive at the 3rd equation? $\endgroup$ – user246836 Sep 8 '15 at 9:01
  • $\begingroup$ @H.Z. Hint: $x^{-4}\log(x^{-4})=\log(x^{-4})\exp(\log(x^{-4}))$. $\endgroup$ – g.kov Sep 8 '15 at 9:47

One may use a slightly geometric argument for this problem. Notice that $\ln(x)=cx^4$ when $f(x)=cx^4-\ln x=0$. Now, $f$ has a vertex, and the vertex occurs where $f'(x)=4cx^3-\frac{1}{x}=0$. Thus, we have the system of equations

\begin{align} f(x)&=cx^4-\ln x=0\\ f'(x)&=4cx^3-\frac{1}{x}=0. \end{align}

Solve the second equation for $x^4$ to get $x^4=\frac{1}{4c}$. Substitute this into the first equation to get $\ln x = \frac{1}{4}$, which implies $x=e^{1/4}$. Finally, substitute this equation for $x$ into $\frac{1}{4}=\ln x=cx^4$ to find that $c=\frac{1}{4e}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.