Solve the following equation:

$$
\log _{\sqrt{x-2}}\left(x^{4}-6 x^{3}+7 x^{2}+9 x-14\right)=8
$$
The expression on the left side of the equation is defined for real numbers if

$$
\text { a) } x>2 ; \quad \text { b) } \sqrt{x-2} \neq 1, \quad \text { i.e., } \quad x \neq 3 ; \quad \text { c) } x^{4}-6 x^{3}+7 x^{2}+9 x-14>0
$$

If the above conditions are satisfied, then our equation is equivalent to the equation

$$
(\sqrt{x-2})^{8}=x^{4}-6 x^{3}+7 x^{2}+9 x-14
$$

Since

$$
x^{4}-6 x^{3}+7 x^{2}+9 x-14=(x-2)\left(x^{3}-4 x^{2}-x+7\right)
$$

rearranging our equation, we get

$$
(x-2)\left[(x-2)^{3}-\left(x^{3}-4 x^{2}-x+7\right)\right]=(x-2)\left(2 x^{2}-13 x+15\right)=0
$$

Due to condition a), \(x-2>0\), so the above equation can only hold if \(2 x^{2}-13 x+15=0\). The roots of this equation are 1.5 and 5. The first one is not a solution to the original equation due to condition a). Therefore, if there is a solution, it can only be \(x=5\). Since \(b\) and \(c\) also hold for this value, it is indeed a solution.

Kóczy T. László, Budapest