5. (3 points) Find all functions continuous on the entire number line that satisfy the identity $5 f(x+y)=f(x) f(y)$ and the condition $f(1)=10$.

Answer: $f(x)=5 \cdot 2^{x}$.
Solution:

First, we prove that \( f(x) \geqslant 0 \). Indeed, \( 5 f(x) = f\left(\frac{x}{2}\right)^{2} \geqslant 0 \).

Substituting \( x = y = 0 \) into the given condition, we get \( 5 f(0) = f(0) \cdot f(0) \), from which \( f(0) = 0 \) or \( f(0) = 5 \). If \( f(0) = 0 \), then \( 5 f(x + 0) = f(x) f(0) \), which implies \( f(x) = 0 \) for any \( x \), which is incorrect. Therefore, \( f(0) = 5 \).

Substituting \( y = 1 \), we get \( 5 f(x + 1) = f(x) \cdot f(1) \), which means \( f(x + 1) = 2 f(x) \). From this, for any natural number \( n \), we find \( f(n) = f(0) 2^{n} = 5 \cdot 2^{n} \).

It is also easy to prove by induction the formula \( f\left(x_{1} + x_{2} + \ldots + x_{n}\right) = \frac{1}{5^{n-1}} f\left(x_{1}\right) f\left(x_{2}\right) \ldots f\left(x_{n}\right) \), from which, in particular, \( f(1) = f\left(\frac{1}{n} + \ldots + \frac{1}{n}\right) = \frac{1}{5^{n-1}} \left(f\left(\frac{1}{n}\right)\right)^{n} \). Thus, \( f\left(\frac{1}{n}\right)^{n} = 2 \cdot 5^{n} \), from which \( f\left(\frac{1}{n}\right) = 5 \cdot 2^{\frac{1}{n}} \) for all natural numbers \( n \).

From this, \( f\left(\frac{m}{n}\right) = f\left(\frac{1}{n} + \ldots + \frac{1}{n}\right) = \frac{1}{5^{m-1}} \left(f\left(\frac{1}{n}\right)\right)^{m} = \frac{1}{5^{m-1}} 5^{m} \cdot 2^{\frac{m}{n}} = 5 \cdot 2^{\frac{m}{n}} \) for natural numbers \( m \) and \( n \).

Now, let's consider negative fractions: \( 5 f(0) = 5 f\left(\frac{m}{n} - \frac{m}{n}\right) = f\left(\frac{m}{n}\right) f\left(-\frac{m}{n}\right) \), from which \( f\left(-\frac{m}{n}\right) = 5 \cdot 2^{-\frac{m}{n}} \).

Thus, the function \( f(x) = 5 \cdot 2^{x} \) for all rational numbers. The function on the right-hand side is also continuous, and if two functions coincide for all rational numbers, they coincide on the entire real line.