$\log x =Cx^4$ 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?
 A: 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. 
A: 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
$$
C=e^{-1/4}.
$$
A: Find the value of $f'(x)$ at $c$, when $f(x) =\log x$, $c=e$.
A: $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_{0}}}\def\Wm{\operatorname{W_{-1}}}\def\e{\mathrm{e}}$
\begin{align} 
\ln(x)&=c\,x^4
\tag{1}\label{1}
.
\end{align}
I'm surprised that the 
Lambert W function
was 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}
$\endgroup$
