5. Solve the equation for all values of the parameter $a$

$$
3 x^{2}+2 a x-a^{2}=\ln \frac{x-a}{2 x}
$$
Solution: If $a=0$, then the equation will take the form $3 x^{2}=\ln \frac{1}{2}$, which has no solutions. If $a \neq 0$, then $x$ does not belong to the interval $[0, a]$ or $[a, 0]$ depending on the sign of the parameter $a$, since otherwise the natural logarithm does not exist. Transform the equation to the form;

$$
4 x^{2}+\ln |2 x|=(x-a)^{2}+\ln |x-a|
$$

Let $f(t)=|t|^{2}+\ln |t|$. Then (*) will take the form $f(2 x)=f(x-a)$. Note that $f(t)$ is an even and strictly increasing function for $t \geq 0$ (since it is the sum of two functions that are increasing for $t \geq 0$). Therefore, $f(2 x)=f(x-a)$ only if $2 x=x-a$ or $-2 x=x-a$. Thus, $x=-a$ or $x=\frac{a}{3}$. But the point $x=\frac{a}{3}$ lies between $a$ and 0, so $x=-a$ is the only solution to the equation.

Answer: When $a=0$, there are no solutions. When $a \neq 0$, the only solution is $x=-a$.

Criteria: Full solution - 14 points. Obtaining an expression of the form (*) - 4 points. If the point $x=\frac{a}{3}$ is not excluded, -4 points. If it is not specified that there are no solutions when $a=0$, -2 points.