Question

Помогите пожалуйста, даю все баллы!
a) log2 (x+4)+log2 (x+2)=3
б) log2*2 x-log2x*3 -4=0

Answer 1

$\displaystyle\bf\\1)\\\\ODZ:\\\\\left \{ {{x+4 > 0} \atop {x+2 > 0}} \right. \\\\\\\left \{ {{x > -4} \atop {x > -2}} \right. \ \ \Rightarrow \ \boxed{ x > -2}\\\\\\\log_{2} (x+4)+\log_{2} (x+2)=3\\\\\log_{2} \Big[(x+4)(x+2)\Big]=3\\\\(x+4)(x+2)=2^{3} \\\\x^{2} +2x+4x+8=8\\\\x^{2} +6x=0\\\\x(x+6)=0\\\\x_{1} =0\\\\x+6=0 \ \ \Rightarrow \ \ x_{2} =-6-neyd\\\\Otvet: \ 0$

$\displaystyle\bf\\2)\\\\ODZ:\\\\x > 0\\\\\\\ log_{2} ^{2} x-\log_{2} x^{3} -4=0\\\\\log_{2} ^{2} x-3\log_{2} x-4=0\\\\\log_{2} x=m\\\\m^{2} -3m-4=0\\\\Teorema \ Vieta:\\\\m_{1} +m_{2} =3\\\\m_{1} \cdot m_{2} =-4\\\\m_{1} =4 \ \ ; \ \ m_{2} =-1\\\\\1) \ \log_{2} x=4\\\\x_{1} =2^{4} =16\\\\2) \ \log_{2} x=-1\\\\x_{2} =2^{-1} =0,5\\\\Otvet: \ 16 \ \ ; \ \ 0,5$