Question

Помогите пожалуйста!

Answer 1

$1)x>0\\\\x^{log_{5}x }=(5^{log_{5}x })^{log_{5}x }=5^{log_{5}^{2}x}\\\\5^{log_{5}^{2}x }+x^{log_{5}x }\geq2\sqrt[4]{5} \\\\5^{log_{5}^{2}x}+5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\2*5^{log_{5}^{2}x}\geq 2\sqrt[4]{5}\\\\5^{log_{5}^{2}x}\geq5^{\frac{1}{4} }\\\\log_{5}^{2} x\geq \frac{1}{4}$

$(log_{5}x-\frac{1}{2})(log_{5}+\frac{1}{2})\geq0\\\\(log_{5}x-log_{5}5^{\frac{1}{2} })(log_{5}x-log_{5}5^{-\frac{1}{2} })\geq0\\\\(5-1)(x-\sqrt{5})(5-1)(x-\frac{1}{\sqrt{5} })\geq0\\\\(x-\sqrt{5})(x-\frac{1}{\sqrt{5} })\geq0\\\\x\in(0;\frac{1}{\sqrt{5} }]\cup[\sqrt{5};+ \infty)$

$2)log_{3}^{2}x+2>3log_{3}x\\\\log_{3}x=m\\\\m^{2}-3m+2>0\\\\(m-1)(m-2)>0\\\\m<1;m>2\\\\log_{3}x<1\\\\x<3\\\\log_{3}x>2\\\\x>9\\\\x\in(0;3)\cup(9;+\infty)$

Ответ :

$x\in(0;\frac{1}{\sqrt{5} }]\cup[\sqrt{5};3)\cup(9;+\infty)$