5. (3 points) Find all functions continuous on the entire number line that satisfy the identity $2 f(x+y)=f(x) f(y)$ and the condition $f(1)=10$.

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

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

Substituting \( x = y = 0 \) into the given condition, we get \( 2 f(0) = f(0) \cdot f(0) \), from which \( f(0) = 0 \) or \( f(0) = 2 \). If \( f(0) = 0 \), then \( 2 f(x + 0) = f(x) f(0) \), which implies \( f(x) = 0 \) for any \( x \), which is incorrect. Therefore, \( f(0) = 2 \).

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

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

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

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

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