$f'(x) = 20x^4 + 3x^2 + 2 > 0$ for all real values $x$, and $f(0) = 1 > 0$, $f(-1) = -6 < 0$, then by the intermediate value theorem there exits one and only one real root in $(-1,0)$ for the equation: $4x^5 + x^3 + 2x + 1 = 0$
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 $\displaystyle c = \left\{\frac{1}{4e}\right\}$
Best Answer
If there are more than two solutions, then $f(x)=e^x-3x$ has at least three zeros. How many zeros must $f'(x)$ have?
To show there are at least two solutions, you might use the Intermediate Value Theorem.