Problem 8. For what values of the parameter $a$ does the equation

$\ln (x-2 a)-3(x-2 a)^{2}+2 a=0$ have a unique root?
Answer: when $a=\frac{\ln 6+1}{4}$.

Solution. Let's make the substitution $t=x-2 a$ and write the equation in the form

$\ln t=3 t^{2}-2 a$. We will plot the graph of the function $y=\ln t$ and the family of parabolas $y=3 t^{2}-2 a$ in the same coordinate system.

![](https://cdn.mathpix.com/cropped/2024_05_06_ec0b1918074dfb35bb6bg-07.jpg?height=548&width=417&top_left_y=780&top_left_x=86)

These curves can have exactly one common point only in the case of tangency. This can be explained as follows: if we consider the graph of $g(t)=$ $\ln t-3 t^{2}+2 a$, it goes to $-\infty$ both at 0 and at $+\infty$. Then, if it has exactly one zero, it must be a local maximum, i.e., a point of tangency with the axis.

Let $t_{0}$ be the abscissa of the point of tangency. Then

$$
\left\{\begin{array}{rl}
\ln t_{0} & = 3 t_{0}^{2} - 2 a_{0} \\
\frac{1}{t_{0}} & = 6 t_{0}
\end{array} \Rightarrow \left\{\begin{array}{rl}
t_{0} & =\frac{1}{\sqrt{6}} \\
\ln \frac{1}{\sqrt{6}} & =\frac{1}{2}-2 a_{0}
\end{array} \Rightarrow a_{0}=\frac{\ln 6+1}{4}\right.\right.
$$