For what positive value of $c$ does the equation $\log(x)=cx^4$ have exactly
one real root?
I think I should find a way to apply IMV and Rolle's theorem to $f(x) = \log(x) – cx^4$. I think I should first find a range of values for $c$ such that the given equation has a solution and then try to find one that gives only a unique solution. I have thought over it but nothing comes to my mind. Perhaps I'm over-complicating it.
Any ideas on how to proceed are appreciated.
Best Answer
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 $\displaystyle c = \left\{\frac{1}{4e}\right\}$