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1+½log₂x=(log₂4+log₂x)/(1-log₂4-log₂x)
log₂x=t
1+½t=(2+t)/(1-2-t)
1+½t=(-2-t)(1+t)
(1+½t)(1+t)=-2-t
1+t+½t²+½t+2+t=0
3+3½t+½t²=0
6+7t+t²=0
Δ=49-24=5²
t₁=(-7+5)/2=-1
t₂=(-7-5)/2=-6
log₂x=-1 czyli x=½
lub
log₂x=-6 czyli x=(½)⁶