2A. It is known that $a^{x}=b^{y}=c^{z}=30^{w}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{w}$, where $a, b, c$ are natural numbers, and $x, y, z, w$ are real numbers. Determine the sum $a+b+c$.
Solution. Note that none of the numbers $a, b, c$ can be 1. If any of the numbers $a, b$ or $c$ were to take the value 1, then $30^{w}=1$, from which $w=0$, which is not possible because according to the problem's condition, $x, y, z, w$ are non-zero real numbers. Now, since $a, b, c>1$, by taking the logarithm of the equation $a^{x}=b^{y}=c^{z}=30^{w}$, we get

$$
x=w \log _{a} 30, y=w \log _{b} 30, z=w \log _{c} 30
$$

From the equation $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{w}$, it follows that

$$
\frac{1}{w \log _{a} 30}+\frac{1}{w \log _{b} 30}+\frac{1}{w \log _{c} 30}=\frac{1}{w}
$$

which means $\log _{30} a+\log _{30} b+\log _{30} c=1$. Therefore, $\log _{30} a b c=1$, so $a b c=30$. Since $a, b, c$ are natural numbers greater than 1, for their product we have $a b c=2 \cdot 3 \cdot 5$, so the numbers $a, b, c$ are $2,3,5$ (in some order), which means $a+b+c=2+3+5=10$.