Let the prime factor decomposition of $x$ be given by $2^a 3^b 5^c p_4^{e_4} \dots p_r^{e_r}$ (with $a, b, c \ge 0$). We conclude that
* $a$ is a multiple of $15$ and is odd,
* $b$ is a multiple of $10$ and $b+1$ is a multiple of $3$,
* $c$ is a multiple of $6$ and $c + 1$ is a multiple of $5$.

The smallest positive integers with these properties are $a = 15$, $b = 20$, $c = 24$. All other exponents in the prime factor decomposition of $x$ must be multiples of $30$, thus we obtain the smallest value for $x$ when all other exponents vanish.

Therefore, the smallest solution is given by $x = 2^{15} \cdot 3^{20} \cdot 5^{24}$.

qed