1) fie [tex]( \sqrt{5}-2)^x=a,a\ \textgreater \ 0 \\ a+ \frac{1}{a}=18 \\ a^{2}-18a+1=0 \\ [/tex]
Δ=324-4=320
[tex] a_{1}= \frac{18-8 \sqrt{5} }{2} =9-4 \sqrt{5} \\ a_{2}=9+ \sqrt{5} \\ [/tex]
revenim la substitutie
[tex]( \sqrt{5}-2)^x=(2+ \sqrt{5})^2 \\ x=-2 \\ [/tex]
