Vijetnam 2007

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

Substitute $y=0$ in (2) we get $$g(x)=g(x)3^{g(0)-1}\forall x\in \mathbb{R}\iff \{\begin{array}{ll}g(x)=0& \forall x\in \mathbb{R}\\ g(0)=1& \end{array}$$ ) 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 $$g(y)=g(0)3^{g(y-1)}\iff g(y)=3^{g(y-1)}\iff 3^{g(y-1)}-g(y)=0,\forall y\in \mathbb{R}.(3)$$ Consider the function $h(t)=3^{t-1}-t$ we get $h^{\prime }(t)=3^{t-1}\ln 3-1$. $$h^{\prime }(t)=0⟺t={\log }_3({\log }_{3}e)+1<1.$$

t	$-\infty$	${\log }_3({\log }_{3}e)+1$	$1$	$+\infty$
$h^{\prime }(t)$	$-$	$0$	$+$	$+$
$h(t)$	(0, $+\infty$)

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, $$g(y)=3^{g(y)-1}⟺\{\begin{array}{l}g(y)=1\\ g(y)=cc=\text{const},0<c<1\end{array}\forall y\in \mathbb{R}(4)$$ 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 $g(-y_0)=\frac{1}{c}\ne c$, which contradicts (4). Therefore $g(y)=1\forall y\in \mathbb{R}$, which implies that $f(x)=1-b^x$.

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