Solve the following equation:

$$
\left(\frac{x}{3}\right)^{3+\log x}=30000
$$
Our equation originates from:

$$
(3+\log x)(\log x-\log 3)=4+\log 3
$$

or

$$
(\log x)^{2}-(\log 3-3) \log x=4+4 \log 3
$$

solving this equation for $(\log x)$:

$$
\log x=\frac{\log 3-3}{2} \pm \frac{\log 3+5}{2}
$$

from which

$$
\log x_{1}=1+\log 3, \log x_{2}=-4
$$

and thus

$$
x_{1}=30, x_{2}=\frac{1}{10000}
$$

(Julius Schlesinger, Budapest, V. district gymnasium.)

The problem was also solved by: Z. Harsányi, P. Heimlich, Gy. Jánosy, S. Pichler, B. Tóth.