Question

Помогите пожалуйста с log

Answer 1

Ответ:

х = 5

Объяснение:

$\displaystyle log_{2}(x-3)+ log_{2}(x)=log_{2}(10)$

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ОДЗ:

$\displaystyle \left \{ {{x > 0} \atop {x-3 > 0}} \right. < = > \left \{ {{x > 0} \atop {x > 3}} \right. < = > x > 3$

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$\displaystyle log_{2}((x-3)*x)=log_{2}(10)$

$\displaystyle x^2-3x=10$

$\displaystyle x^2-3x-10=0$

$\displaystyle D = (-3)^2-4*1*(-10) = 9+40 = 49 = 7^2$

$\displaystyle x_{1}=\frac{3+7}{2*1}=\frac{10}{2} = 5$

$\displaystyle x_{2}=\frac{3-7}{2*1}=-\frac{4}{2} = -2$ - не подходит по ОДЗ

Answer 2

$\displaystyle\bf\\ODZ:\\\\\left \{ {{x-3 > 0} \atop {x > 0}} \right. \\\\\left \{ {{x > 3} \atop {x > 0}} \right. \ \ \Rightarrow \ \ x > 3\\\\\\\log_{2} (x-3)+\log_{2} x=\log_{2} 10\\\\\log_{2} \Big[(x-3)\cdot x\Big]=\log_{2} 10\\\\(x-3)\cdot x=10\\\\x^{2} -3x-10=0\\\\Teorema \ Vieta:\\x_{1}+x_{2} =3\\\\x_{1} \cdot x_{2} =-10\\\\x_{1} =5 \ \ ; \ \ x_{2} =-2 < 3-neyd\\\\Otvet: \ 5$