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

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

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}