$\log x =Cx^4$ has only one root. Find C
$\log x =Cx^4$ has only one root. Find C.
I don't know how to solve this problem. Do you take derivative on both sides?
I am thinking C equals 0. Am I correct on that?
Solution 1:
Generally a function has a root, not an equation. I am assuming here that the 'root' is really a solution to the equation.
Plot $\log x$ and $cx^4$ simultaneously for various values of $c$; you will notice the following:
- For large positive $c$, there are no intersection points.
- At some critical positive $c$, the two graphs are tangent with one intersection point.
- For positive $c$ smaller than $c_{crit}$, there are two intersections (one near 1, and one which tends to infinity).
- At $c=0$, there is only the root left at $x=1$ -- the other root has escaped to infinity.
- For negative $c$, there is always precisely one solution for positive $x$, tending to 0 as $c \to -\infty$.
To find the value of $c_{crit}$, note that the graphs must be tangent and differentiate to find that $$ \log x = c_{crit}x^4 ~~\text{and}~~ \frac 1 x = 4c_{crit}x^3 ~~\text{so that after some algebra,}~~ x^4 = \frac 1 {4c}, c_{crit} = \frac 1 {4e} . $$ Hence there is precisely one solution for $c=0,c=\frac 1 {4e}$ with values $x = 1,e^{1/4}$ respectively. If negative $c$ are permitted, then all negative $c$ work too, with precise root given by a Lambert-W expression.
Solution 2:
You can examine the function $$ f(x)=\log x-Cx^4 $$ which is defined for $x>0$. We have $\lim_{x\to0}f(x)=-\infty$ and $$ \lim_{x\to\infty}f(x)= \begin{cases} \infty & \text{if $C \le 0$},\\ -\infty & \text{if $C > 0$}. \end{cases} $$ Compute the derivative $$ f'(x)=\frac{1}{x}-4Cx^3=\frac{1-Cx^4}{x} $$ from which we deduce that the derivative is everywhere positive when $C\le 0$. So, for $C\le 0$ the function $f$ is strictly increasing and the equation $f(x)=0$ has a unique solution.
Let's now look at the $C>0$ case, where the derivative vanishes only at $C^{-1/4}$ which therefore is where the function $f$ attains its maximum.
If the maximum value is positive, the equation $f(x)=0$ has two solutions, if it's negative there will be no solution, if the maximum value is $0$ the solution is unique.
Since $$ f(C^{-1/4})=-\frac{1}{4}\log C - 1 $$ we can conclude that the value we're looking for is $$ C=e^{-1/4}. $$