For which value of $a$ does the equation

$$
10^{x^{2}+4 x+1}=a
$$

have real solutions?
First of all, we see that $a$ must be a positive number, because every real power of 10 is a positive number. If we take the logarithm of both sides of our equation, we get:

$$
x^{2}+4 x+1=\log a
$$

This equation has a real solution if

$$
12+4 \log a \geqq 0
$$

from which

$$
3+\log a \geqq 0
$$

and thus

$$
\log a \geqq-3
$$

or

$$
a \geqq \frac{1}{1000}
$$

(Czúcz Ambró, Eger.)