b)
log(x-5)-log2=\frac{1}{2}log(3x-20)
załozenia:
x-5>0\to x>5
i
3x-20>0\to 3x>20 \to x>\frac{20}{3}
2log(x-5)-2log2=log(3x-20)
log(x-5)^2-log4=log(3x-20)
log\frac{(x-5)^2}{4}=log(3x-20)
\frac{(x-5)^2}{4}=3x-20
x^2-10x+25=4(3x-20)
x^2-10x+25=12x-80
x^2-22x+105=0
a=1, b=-22, c=105
\Delta=b^2-4ac=484-420=64
\sqrt\Delta=8
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{22-8}{2}=7
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{22+8}{2}=15