$\log x =Cx^4$ 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?
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$\begingroup$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.
$\endgroup$ 7 $\begingroup$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 $$ C=e^{-1/4}. $$
$\endgroup$ $\begingroup$Find the value of $f'(x)$ at $c$, when $f(x) =\log x$, $c=e$.
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$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_{0}}}\def\Wm{\operatorname{W_{-1}}}\def\e{\mathrm{e}}$
\begin{align} \ln(x)&=c\,x^4 \tag{1}\label{1} . \end{align}
I'm surprised that the Lambert W functionwas only briefly mentioned in one answer, since it's purpose is to greatly simplify the solutions of exactly this king of equations.
We just need to transform \eqref{1} to the form $u\exp(u)=v$, apply Lambert W function to both sides of it and check the argument of $\W(v)$to find out the number of the real solutions:
\begin{align} 4\,\ln(x) &=4\,c\,x^4 \tag{2}\label{2} ,\\ \tfrac1{x^4}\,\ln(x^4) &=4\,c \tag{3}\label{3} ,\\ -\tfrac1{x^4}\,\ln(x^4) &=-4\,c \tag{4}\label{4} ,\\ \tfrac1{x^4}\,\ln(\tfrac1{x^4}) &=-4\,c \tag{5}\label{5} ,\\ \ln(\tfrac1{x^4})\,\exp(\ln(\tfrac1{x^4})) &=-4\,c \quad (\text{ this is the sought form } u\exp(u)=v ,\quad u=\ln(\tfrac1{x^4}),\quad v=-4\,c) \tag{6}\label{6} \end{align}
\begin{align} \W\left(\ln(\tfrac1{x^4})\,\exp(\ln(\tfrac1{x^4}))\right) &=\W(-4\,c) \tag{7}\label{7} ,\\ \ln(\tfrac1{x^4}) &=\W(-4\,c) \tag{8}\label{8} ,\\ \tfrac1{x^4} &=\exp(\W(-4\,c)) \tag{9}\label{9} ,\\ x^4 &=\exp(-\W(-4\,c)) \tag{10}\label{10} ,\\ x &=\pm \exp(-\tfrac14\,\W(-4\,c)) \tag{11}\label{11} . \end{align}
At this point we need to recall that $x\le0$ are not solutions to \eqref{1} (spurious roots were introduced in step \eqref{3}), so we need to check only positive solutions
\begin{align} x &=\exp(-\tfrac14\,\W(-4\,c)) \tag{12}\label{12} . \end{align}
In \eqref{12} the argument of $\W(-4\,c)$ is $-4\,c$, hence
there are no real solutions for $c>\tfrac1{4\e}$, there are one real solution \begin{align} x &=\exp(-\tfrac14\,\Wp(-4\,c))\quad \text{for}\quad c\le0 \\ \text{and}\quad x &=\exp(-\tfrac14\,\Wp(-\tfrac1{\e}))=\exp(-\tfrac14\,\Wm(-\tfrac1{\e})) =\exp(\tfrac14) \quad \text{for}\quad c=\tfrac1{4\e} , \end{align}
and there are two real solutions
\begin{align} x_0 &=\exp(-\tfrac14\,\Wp(-4\,c)) ,\\ \text{and}\quad x_{-1} &=\exp(-\tfrac14\,\Wm(-4\,c)) \quad \text{for}\quad c \in(0,\tfrac1{4\e}) . \end{align}
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