, 2007

The given equation system is equivalent to the following equation system
$$f(x+y) + b^{x+y} = (f(x)+b^x)3^{b^x+f(y)-1} \quad \forall x,y \in \mathbb{R} \quad (1)$$
Let $g(x) = f(x) + b^x$. Then $(1) \Leftrightarrow g(x+y) = g(x)3^{g(y)-1} \quad \forall x, y \in \mathbb{R} \quad (2)$

Substitute $y = 0$ in (2) we get
$$g(x) = g(x)3^{g(0)-1} \quad \forall x \in \mathbb{R} \quad \Leftrightarrow \quad \begin{cases} g(x) = 0 & \forall x \in \mathbb{R} \\ g(0) = 1 \end{cases}$$
*) If $g(x) = 0$ $\forall x \in \mathbb{R}$ then $f(x) = -b^x$.

*) If $g(0)=1$, then by putting $x=0$ in (2) we get
$$g(y) = g(0)3^{g(y-1)} \Leftrightarrow g(y) = 3^{g(y-1)} \Leftrightarrow 3^{g(y-1)} - g(y) = 0, \forall y \in \mathbb{R}. \quad (3)$$
Consider the function $h(t) = 3^{t-1} - t$ we get $h'(t) = 3^{t-1} \ln 3 - 1$.
$$h'(t) = 0 \iff t = \log_3(\log_3 e) + 1 < 1.$$

From the table we easily see that the equation $h(t) = 0$ has two roots $t_1 = 1$ and $t_2 = c$ with $0 < c < 1$. Thus,
$$g(y) = 3^{g(y)-1} \iff \begin{cases} g(y) = 1 \\ g(y) = c \quad c = \text{const}, 0 < c < 1 \end{cases} \quad \forall y \in \mathbb{R} \quad (4)$$
Suppose that there exists $y_0 \in \mathbb{R}$ such that $g(y_0) = c$. Then
$$1 = g(0) = g(y_0 - y_0) = g(-y_0) \cdot 3^{g(y_0)-1} = c \cdot g(-y_0).$$
This implies that $g(-y_0) = \frac{1}{c} \neq c$, which contradicts (4). Therefore $g(y) = 1 \quad \forall y \in \mathbb{R}$, which implies that $f(x) = 1 - b^x$.

Thus, there are two functions $f(x) = -b^x$ and $f(x) = 1 - b^x$.