Reguli de calcul cu puteri:
[tex]9^{x}+2\cdot 3^x=99\\\\(3^2)^{x}+2\cdot 3^x=3^2\cdot 11\\\\3^{2x}+2\cdot 3^x=3^2\cdot 11[/tex]
Notam [tex]3^x=t[/tex]
t²+2t=99
t²+2t-99=0
Δ=4+396=400
[tex]t_1=\frac{-2+20}{2} =9\\\\3^x=9\\\\x=2[/tex]
[tex]t_2=\frac{-2-20}{2} =-11 < 0\ NU[/tex]
Un exercitiu cu calcul puteri gasesti aici: https://brainly.ro/tema/1043829
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