FORM GR1268 Question 30

Question:
For what positive value of \(c\) does the equation \(\log x=cx^4\) have exactly one real solution for \(x\)?

(A) \(\displaystyle\frac{1}{4e}\)     (B) \(\displaystyle\frac{1}{4e^4}\)     (C) \(\displaystyle\frac{e^4}{4}\)     (D) \(\displaystyle\frac{4}{e^{1/4}}\)     (E) \(\displaystyle 4e^{1/4}\)

Answer:
(A)

Answer Key:
Depicted below are graphs of \(f(x)=\log x\) and \(g(x)=cx^4\) with \(c=1\).

We want to adjust \(c\) so that two curves touch.

Suppose the two curves touch at \(x=x_0\).

Then \(f'(x_0)=g'(x_0)\), so
\[\begin{align}
\frac{1}{x_0}&=4cx_0^3\\
1&=4cx_0^4\\
x_0^4&=\frac{1}{4c}\\
x_0&=\left(\frac{1}{4c}\right)^{\frac{1}{4}}.
\end{align}\]
We also have \(f(x_0)=g(x_0)\). Thus,
\[\begin{align}
\log x_0&=cx_0^4\\
\log\left(\frac{1}{4c}\right)^{\frac{1}{4}}&=c\cdot\frac{1}{4c}\\
\frac{1}{4}\log\frac{1}{4c}&=\frac{1}{4}\\
\log\frac{1}{4c}&=1\\
\frac{1}{4c}&=e\\
4c&=\frac{1}{e}\\
c&=\frac{1}{4e}.
\end{align}\]