notam
[tex] (2+ \sqrt{3} )^{x}=t;t\ \textgreater \ 0 \\ t- \frac{1}{t} =2 \sqrt{3} \\ t^{2}-2 \sqrt{3}t-1=0 [/tex]
Δ=12+4=16
[tex] t_{1}= \frac{2 \sqrt{3}-4 }{2}= \sqrt{3}-2\ \textless \ 0 \\ t_{2}= \frac{2 \sqrt{3}+4 }{2}=2+ \sqrt{3} [/tex]
revenim la substitutie
[tex](2+ \sqrt{3})^x =2+ \sqrt{3} \\ x=1[/tex]
