Question

помогите ❗❗❗❗
логарифмы​

Answer 1

$1. a) \log_3(x-2)+\log_3(x+6)=2; OD3: \left \{ {{x-2>0} \atop {x+6>0}} \right. \Rightarrow x > 2\\\\\log_3((x-2)(x+6))=\log_39\\\\(x-2)(x+6)=9,\\\\x^2+4x-12-9=0;\\\\x^2+4x-21=0.\\\\\left \{ {{x_1+x_2=-4,} \atop {x_1x_2=-21}} \right. \Rightarrow x_1=-7\notin OD3, x_2=3.$

ОТВЕТ: 3.

$b) \log_3^2x+log_3x^2=8; OD3: x>0\\\\\log_3^2x+2\log_3x-8=0. \\\\\log_3x=t,\\\\t^2+2t-8=0\Rightarrow t_1=2, t_2=-4.\\\\\left \ [ {{\log_3x=2} \atop {\log_3x=-4;}} \right. \left \ [ {{x=9} \atop {x=\frac{1}{81} }} \right.$

ОТВЕТ: $\frac{1}{81} ; 9$.

$2. \log_{\frac{1}{3}}(x-1) \geq -2; OD3: x - 1 > 0\Rightarrow x>1.\\x-1\leq (\frac{1}{3} )^{-2},\\x-1\leq 9; \\x \leq 10$

ОТВЕТ: (1; 10].

$3. \frac{1}{5-\lg x}+\frac{2}{1+\lg x}=1.\\ OD3: \left \{ {{\lg x\neq 5} \atop {\lg x\neq -1}} \right. \Rightarrow x\neq 0,1; 100000.\\\\\lg x = t \neq -1;5.\\\\\frac{1}{5-t}+\frac{2}{1+t}=1,\\\\ \frac{1+t+10-2t}{(5-t)(1+t)}=1,\\\\ -t+11=5+4t-t^2,\\\\t^2-5t+6=0\Rightarrow x_1=2,x_2=3.\\\left \ [ {{\lg x =2} \atop {\lg x=3}} \right. \Rightarrow \left \ [ {{x=100} \atop {x=1000}} \right.$

ОТВЕТ: 100; 1000.

$4. \log_3x-\log_x3\leq 1,5. OD3: \left \{ {{x>0} \atop {x\neq 1}} \right. \\\\\log_3x-\frac{1}{\log_3x}-1,5\leq 0. \\\\ \log_3x=t\neq 1.\\\\t-\frac{1}{t}-1,5\leq 0|\cdot2t\\\\ 2t^2-3t-1\leq 0;\\\\D=(-3)^2-4\cdot4\cdot(-1)=9+16=25,\\\\t_{1,2}=\frac{3\pm5}{4}; t_1=2, t_2=-0,5.\\\\ t\in[-0,5;2].\\\\ -0,5\leq \log_3x\leq 2\\\\3^{-0,5}\leq x\leq 3^2\\\\\frac{\sqrt3}{3} \leq x\leq 9$

С учетом условия x > 0, x ≠ 1, x ∈ (0; √3/3] ∪ (1; 3].

ОТВЕТ: (0; √3/3] ∪ (1; 3]