Question

Решите, пожалуйста, очень нужно

Answer 1

1) ОДЗ : x > 0

$log_{2}^{2}(4x)+log_{2}\frac{x^{2}}{8}=8\\\\(log_{2}(4x))^{2}+log_{2}x^{2}-log_{2}8=8\\\\(log_{2}4+log_{2}x)^{2}+2log_{2}x-log_{2}2^{3}=8\\\\(2+log_{2}x)^{2}+2log_{2} x-3-8=0\\\\4+4log_{2}x+log_{2}^{2}x+2log_{2}x -11=0\\\\log_{2}^{2}x+6log_{2}x-7=0\\\\log_{2}x_{1} =-7\\\\x_{1} =\frac{1}{128}\\\\log_{2}x_{2}=1\\\\x_{2}=2$

ОДЗ :

1) x + 1 > 0

x > - 1

2) x + 1 ≠ 1

x ≠ 0

$2 + 3log_{x+1}3\geq log_{3}(x+1)\\\\\frac{3}{log_{3}(x+1)}-log_{3}(x+1)+2\geq0\\\\log_{3}(x+1)=m\\\\\frac{3}{m}-m+2\geq0\\\\\left \{ {{-m^{2}+2m+3\geq 0 } \atop {m\neq0 }} \right.\\\\m^{2}-2m-3\leq0\\\\(m-3)(m+1)\leq0$

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________[- 1]_________[3]________

$1)log_{3}(x+1)\geq-1\\\\x+1\geq\frac{1}{3}\\\\x\geq-\frac{2}{3}\\\\2)log_{3} (x+1)\leq3\\\\x+1\leq27\\\\x\leq 26\\\\x\in[-\frac{2}{3};0)\cup(0;26]$