Task B-4.6. Determine the product of the solutions of the equation

$$
x^{\log _{2011} x} \cdot \sqrt{2011}=x^{2011}
$$
Solution. Taking the logarithm to base 2011:

$$
\begin{aligned}
x^{\log _{2011} x} \cdot \sqrt{2011} & =x^{2011} \\
\log _{2011} x \cdot \log _{2011} x+\log _{2011} \sqrt{2011} & =\log _{2011} x^{2011} \\
\log _{2011}^{2} x+\frac{1}{2} & =2011 \cdot \log _{2011} x
\end{aligned}
$$

The equation makes sense only for $x>0$.

Let's introduce the substitution $\log _{2011} x=a$ and we get

$$
a^{2}-2011 a+\frac{1}{2}=0
$$

The solutions of this equation satisfy Vieta's formulas, so we have:

$$
\begin{array}{ll}
a_{1}+a_{2}=2011 \\
\log _{2011} x=a_{1} \quad \Rightarrow \quad x_{1}=2011^{a_{1}} \\
\log _{2011} x=a_{2} \quad \Rightarrow \quad x_{2}=2011^{a_{2}}
\end{array}
$$

It follows that

$$
x_{1} \cdot x_{2}=2011^{a_{1}} \cdot 2011^{a_{2}}=2011^{a_{1}+a_{2}}=2011^{2011}
$$