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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.

user246836
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5 Answers5

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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 $\displaystyle c = \left\{\frac{1}{4e}\right\}$

Tahlor
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juantheron
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  • $\ln(x) = 0\Rightarrow x = e^{0} = 1$ and $\displaystyle c= \frac{1}{2e}$ – juantheron Sep 08 '15 at 07:39
  • Put $x=1$ in $\ln(x)$ – juantheron Sep 08 '15 at 07:41
  • Oops, right. Fortunately the problem has excluded non-positive solutions for $c$. XD – user246836 Sep 08 '15 at 07:43
  • Sorry HZ I have edited by solution, You are Right, Ans $\displaystyle = \frac{1}{4e}$ – juantheron Sep 08 '15 at 07:45
  • 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? – rosterherik Oct 26 '18 at 22:10
  • Call the functions $f$ and $g$. I think it is because in the case that they only intersect once, we have that $f \geq g$, so $d(f,g) = f - g.$ Then, the minimum of this is at the point of intersection, $x_0$, so $\frac{d}{dx}d(f,g)(x_0) = 0 \iff f'(x_0) = g'(x_0)$ – CalabiYau Aug 04 '22 at 22:43
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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}

Adriano
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1

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 $c={1\over 4e}$.

uniquesolution
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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 $c=\tfrac14\exp(-1)$.

g.kov
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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 $c=\frac{1}{4e}$.

rosterherik
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