Question

Помогите срочно Алгебра 10 логарифмические уравнения 100 баллов, пожалуйста!

Answer 1

$1) 3^{z}=11\\ log_{3}3^{z}=log_{3}11\\ z=log_{3}11\\\\2)log_{0,8}(6+2x)= 1\\\\\left \{ {{6+2x>0} \atop {6+2x=0,8}} \right.\\\\\left \{ {{2x>-6} \atop {2x=-5,2}} \right.\\\\\left \{ {{x>-3} \atop {x=-2,6}} \right.$

Ответ : - 2,6

3) ОДЗ :

1) 7x - 1 > 0   ⇒   x > 1/7

2) x > 0

$log_{17}(7x-1)-log_{17}x= 0\\\\log_{17}\frac{7x-1}{x}=0\\\\\frac{7x-1}{x}=1\\\\7x-1=x\\\\7x-x=1\\\\6x=1\\\\x=\frac{1}{6}$

Ответ : 1/6

$4)log_{2}^{2}x=log_{2}x^{3}-2\\\\x>0\\\\log_{2}^{2}x-3log_{2}x+2=0\\\\log_{2}x=m\\\\m^{2}-3m+2=0\\\\m_{1} =1\\\\m_{2}=2\\\\log_{2}x=1\\\\x_{1}=2\\\\log_{2}x=2\\\\x_{2} =4$

5) ОДЗ :

1) z > 0

2) z ≠1

$log_{0,1z}z+log_{0,25z}z=0\\\\\frac{lgz}{lg0,1z}+\frac{lgz}{lg0,25z} =0\\\\\frac{lgz}{lg0,1+lgz}+\frac{lgz}{lg0,25+lgz}=0\\\\\frac{lgz}{lgz-lg10}+\frac{lgz}{lgz-lg8}=0\\\\lgz=m\\\\\frac{m}{m-1}+\frac{m}{m-lg8}=0\\\\m^{2}-lg8m+m^{2}-m=0\\\\m\neq1;m \neq lg8\\\\2m^{2}-(1-lg8)m=0\\\\m(2m-(1-lg8))=0\\\\m_{1}=0\\\\2m=1-lg8\\\\m_{2} =\frac{1-lg8}{2}=\frac{lg1,25}{2} \\\\lgz=0\\\\z_{1}=1\\\\lgz=\frac{lg1,25}{2}$

$z_{2}=(10^{lg\frac{5}{4} })^{\frac{1}{2} }=\sqrt{\frac{5}{4} }=0,5\sqrt{5}$