2. Solve the equation $\left(\frac{x}{243}\right)^{\log _{2}\left(\frac{9 x}{4}\right)}=\frac{729}{x^{4}}$.
Answer: $x=\frac{243}{4}, x=\frac{1}{9}$.

Solution. By taking the logarithm with base 2, we obtain an equation equivalent to the original:

$$
\log _{2}\left(\frac{x}{243}\right) \cdot \log _{2}\left(\frac{9 x}{4}\right)=\log _{2}\left(\frac{729}{x^{4}}\right) \Leftrightarrow\left(\log _{2} x-5 \log _{2} 3\right)\left(-2+\log _{2} x+2 \log _{2} 3\right)=6 \log _{2} 3-4 \log _{2} x
$$

Let $\log _{2} x=y, \log _{2} 3=a$. After expanding the brackets and combining like terms, the equation becomes $y^{2}-(3 a-2) y+4 a-10 a^{2}=0$. We solve the quadratic equation for $y$:

$$
D=(3 a-2)^{2}-4\left(4 a-10 a^{2}\right)=49 a^{2}-28 a+4=(7 a-2)^{2}, y=\frac{3 a-2 \pm(7 a-2)}{2} \Leftrightarrow\left[\begin{array}{l}
y=5 a-2 \\
y=-2 a
\end{array}\right.
$$

We find $x$: if $y=-2 a$, then $\log _{2} x=-2 \log _{2} 3$ and $x=\frac{1}{9}$; if $y=5 a-2$, then $\log _{2} x=5 \log _{2} 3-2$ and $x=\frac{243}{4}$.