Question

Log^2 2 x + log 2 x =80

Answer 1

Log₂² x + 2log₂ x= 80
x>0
log₂ x = t
t² + 2t - 80 = 0
D=(4+320)=324=18²
t₁₂= (-2+-18)/2 = 8 -10
log₂ x = 8  x=2⁸
log₂ x = -10 x=2⁻¹⁰
ответ 2⁸  2⁻¹⁰