Răspuns:
log₂(x+3) + log₂x = 2;
Conform teoremei adunarii logaritmilor ( logₐx+logₐy = logₐxy) :
log₂(x+3) + log₂x = log₂(x²+3x) = 2 ; x>0
Dar 2 = log₂4 => log₂(x²+3x) = log₂4 => x²+3x=4 => x²+3x-4=0
x²+3x-4=0 :
Δ=9+16=25 ; √Δ=5 => x= -3±√Δ/2a => x₁= -8/2 = -4 => fals pt ca x>0.
x₂=2/2=1 ; x₂>0 => x₂= x = 1