[tex]\it 2+\sqrt3\ \ \c{s}i\ \ 2-\sqrt3\ sunt\ numere\ ira\c{\it t}ionale\ conjugate,\ \ deci:\\ \\ 2-\sqrt3=\dfrac{1}{2+\sqrt3} \Rightarrow (2-\sqrt3)^x=\Big(\dfrac{1}{2+\sqrt3}\Big)^x \Rightarrow (2-\sqrt3)^x=\dfrac{1}{(2+\sqrt3)^x}[/tex]
[tex]\it Not\breve am\ (2+\sqrt3)^x=t,\ t>0,\ iar\ ecua\c{\it t}ia\ dat\breve a\ devine:\\ \\ t+\dfrac{1}{t}=14 \Rightarrow t^2+14t-1=0 \Rightarrow t_{1,2}=7\pm4\sqrt3=(2\pm\sqrt3)^2[/tex]
Revenim asupra notației și vom avea:
[tex]\it (2+\sqrt3)^x=(2+\sqrt3)^2 \Rightarrow x=2\\ \\ (2+\sqrt3)^x=(2-\sqrt3)^2\Rightarrow (2+\sqrt3)^x=(2+\sqrt3)^{-2}\Rightarrow x=-2\\ \\ S=\{-2,\ \ 2\}[/tex]
