Let's find the sum of the following series:

$$
S_{n}=x^{\lg a_{1}}+x^{\lg a_{2}}+\ldots+x^{\lg a_{n}}
$$

where $x$ is a positive number different from 1, and $a_{1}, a_{2}, \ldots, a_{n}$ are distinct positive numbers that form a geometric sequence.
Let the sequence $a_{1}, a_{2}, \ldots, a_{n}$ have a common ratio $q$; by assumption, $q>0$. Furthermore, let $\lg a_{1}=b$ and $\lg q=d$; by assumption, $q \neq 1$ and thus $d \neq 0$. Then for $k=1,2, \ldots, n$,

$$
\lg a_{k}=\lg \left(a_{1} q^{k-1}\right)=\lg a_{1}+(k-1) \lg q=b+(k-1) d
$$

and thus

$$
S_{n}=x^{b}+x^{b+d}+x^{b+2 d}+\ldots+x^{b+(n-1) d}=x^{b}\left(1+x^{d}+x^{2 d}+\ldots+x^{(n-1) d}\right)
$$

The expression in parentheses is the sum of a geometric series. Since its common ratio $x^{d}$ is different from 1 by assumption, we can use the known sum formula:

$$
S_{n}=x^{b} \frac{x^{n d}-1}{x^{d}-1}
$$

Finally, returning to the given data, $d=\lg q=\lg a_{2}-\lg a_{1}$, and thus

$$
S_{n}=x^{\lg a_{1}} \frac{x^{n}\left(\lg a_{2}-\lg a_{1}\right)-1}{x^{\lg a_{2}-\lg a_{1}}-1}
$$

Eva Porzsolt (Nyíregyháza, Kölcsey F. g. III. o. t.)

Remarks: Omitting either the condition $x \neq 1$ or the requirement that $a_{1}, a_{2}, \ldots, a_{n}$ are distinct, the sum formula cannot be used, but $S_{n}$ can be expressed much more simply because all its terms are equal. For $x=1$, regardless of $a_{1}, a_{2}, \ldots, a_{n}$, $S_{n}=n\left(a_{1}, a_{2}, \ldots, a_{n}\right.$ positivity is still used here because $\lg a_{1}, \lg a_{2}, \ldots, \lg a_{n}$ only exist in this case); for $x \neq 1, a_{1}, a_{2}, \ldots, a_{n}(>0)$, $S_{n}=n x^{\lg a_{1}}$.

The positivity of $x$ was seemingly not used. The problem only required $x>0$ to ensure that all terms of $S_{n}$ always have meaning, even if irrational numbers appear in the exponents. During the following transformation, the existence of $\lg x$ will be sharply used, as $x>0$:

$$
x^{\lg a_{1}}=\left(10^{\lg x}\right)^{\lg a_{1}}=\left(10^{\lg a_{1}}\right)^{\lg x}=a_{1}^{\lg x}
$$

Similarly, $x^{\lg a_{2}}=a_{2}^{\lg x}$, and thus

$$
S_{n}=a_{1}^{\lg x}+a_{2}^{\lg x}+\ldots+a_{n}^{\lg x}=a_{1}^{\lg x} \frac{\left(\frac{a_{2}}{a_{1}}\right)^{n \lg x}-1}{\left(\frac{a_{2}}{a_{1}}\right)^{\lg x}-1}=\frac{a_{2}^{n \lg x}-a_{1}^{n \lg x}}{a_{1}^{n \lg x}\left(a_{2}^{\lg x}-a_{1}^{\lg x}\right)}
$$