Problem 3. (4 points) Given a positive number $a$, for which the inequality $10<a^{x}<100$ has exactly 5 solutions in natural numbers. How many natural number solutions can the inequality $100<a^{x}<1000$ have?
Solution. Let $a=10^{b}$. The inequality $10<a^{x}<100$ can be written as $10<10^{b x}<100$, i.e., $1<b x<2$. Similarly, $100<a^{x}<1000$ is equivalent to $2<b x<3$. If $n$ is the smallest natural solution to $1<b x<2$, then

$$
b(n-1)<1<b n \quad \text { and } \quad b(n+4)<2<b(n+5) .
$$

By doubling the first and adding the first and second inequalities, we get

$$
b(2 n-2)<2<b(2 n) ; \quad \text { and } \quad b(2 n+3)<3<b(2 n+5) .
$$

From the last two inequalities, it follows that the inequality $2<b x<3$ can have 4, 5, or 6 natural solutions.

All three cases are possible: for $b=\frac{5}{23}$, the solutions to the first inequality are $5,6,7,8$ and 9, and to the second: $10,11,12$ and 13; for $b=\frac{5}{26}$, the solutions to the first inequality are $6,7,8,9$ and 10, and to the second: $11,12,13,14$ and $15;$ for $b=\frac{5}{27}$, the solutions to the first inequality are $6,7,8,9$ and 10, and to the second: 11, $12,13,14,15$ and 16.