75th Romanian Mathematical Olympiad

There are no numbers $x\in [1,\infty )$ with the property in the statement: if $x\ge 1$, then $[x]\ge 1$ and, since $\{x\}\in [0,1)$, we obtain $$4=3^x+3^{[x]}+3^{\{x\}}\ge 3+3+1=7,$$ contradiction.

There are no numbers $x\in (-\infty ,-1)$ with the property in the statement: $x<-1$, then $[x]\le -2$ and, since $\{x\}\in [0,1)$, we obtain $$4=3^x+3^{[x]}+3^{\{x\}}<3^{-1}+3^{-2}+3^1=3\frac{4}{9},$$ contradiction.

If $x\in [0,1)$, then $[x]=0$, $\{x\}=x$, so the given equality becomes $$3^x+1+3^x=4\iff 2\cdot 3^x=3.$$ We obtain the solution $x_1=1-{\log }_{3}2\in [0,1)$.

If $x\in [-1,0)$, then $[x]=-1$, $\{x\}=x+1$, so the given equality becomes $$3^x+\frac{1}{3}+3^{x+1}=4\iff 4\cdot 3^x=\frac{11}{3}.$$ We obtain the solution $x_2={\log }_3\frac{11}{12}\in [-1,0)$.