Question

ajutati.ma, va rog! log(1-x)=3-log(3-x) logaritmii sunt in baza 2. Multumesc

Answer 1

log(2) (1-x) =log(2) (2³) – log(2)(3-x)

1-x>0 si 3-x>0 => x<1

log(2) (1-x) =log(2)[ 8/ log(2)(3-x)]

1-x=8/(3-x)

(1-x)(3-x)=8

3-4x+x²=8

x²-4x-5=0

Δ=16+20=36

x1=(4+6)/2=5, nu convine

x2=(4-6)/2=-1

x= - 1