Problem 8. For what values of the parameter $a$ does the equation $\ln (x+a)-4(x+a)^{2}+a=0$ have a unique root 3
Answer: when $a=\frac{3 \ln 2+1}{2}$.

Solution. Let's make the substitution $t=x+a$ and write the equation in the form $\ln t=4 t^{2}-a$. We will plot the graph of the function $y=\ln t$ and the family of parabolas $y=4 t^{2}-a$ in the same coordinate system.

![](https://cdn.mathpix.com/cropped/2024_05_06_ec0b1918074dfb35bb6bg-04.jpg?height=551&width=417&top_left_y=1078&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 $g(t)=$ $\ln t-4 t^{2}+a$, 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

$$
\left\{\begin{array}{rl}
\ln t_{0} & = 4 t_{0}^{2} - a_{0} \\
\frac{1}{t_{0}} & = 8 t_{0}
\end{array} \Rightarrow \left\{\begin{array}{rl}
t_{0} & =\frac{1}{2 \sqrt{2}} \\
\ln \frac{1}{2 \sqrt{2}} & =\frac{1}{2}-a_{0}
\end{array} \Rightarrow a_{0}=\frac{3 \ln 2+1}{2}\right.\right.
$$