a) 4^x + 2^x+1 = 80
(2^2)^x + 2^x • 2 = 80
(2^x)^2 + 2^x • 2 = 80
inlocuim 2^x cu y
y^2 + 2y = 80
y^2 + 2y - 80 = 0
y^2 + 10y -8y -80 = 0
y ( y + 10 ) -8 ( y - 10 ) = 0
( y + 10 ) • ( y - 8 ) = 0
=> y + 10 = 0
=> y - 8 = 0
=> y = -10
=> y = 8
=> 2^x = -10 => puterea nu da cu minus
=> 2^x = 8 => 2^x = 2^3 => x = 3
