Question

rezolvati ecuatia:9^x+2*3^x=99(^=la puterea)

Answer 1

 9^x  = (3 ^x   ) ² = t² 
ec . exp. devine rez. ec. de gr. II 
t² + 2·t  - 99 =0 
Δ = 4 - 4 · ( - 99 ) = 4 + 4 · 99 = 4 ( 1 + 99 ) = 4 · 100 
√Δ = 20 
t = ( -2 -20 ) / 2 = -  22 /2 = -11       FALS 
t =( - 2 + 20 ) /2 =  9             ⇒ 9 =3^x   = 3²  ⇒   solutia unica x = 2

Answer 2

$9^x+2*3^x=99 \\ 9^x+2*3^x-99=99-99 \\ 9^x+2*3^x-99=0 \\ (3^x-9)(3^x+11)=0 \\ 3^x-9=0 \\ 3^x-9+9=0+9 \\ 3^x=9 \\ 3^x=3^2 \\ x=2 \\ 3^x+11=0 \\ 3^x+11-11=0-11 \\ 3^x=-11 \\ x=2$