Question

log3(2x-1)+log3(x+3)=2
log0,2x(меньше или равно)-3

Answer 1

#1
$log_{_{3}}(2x-1) + log_{_{3}}(x+3) = 2 \\ \\ log_{_{3}}((2x-1)(x+3)) = 2 \\ \\ log_{_{3}}(2x^{2}+5x-3) = 2 \\ \\ 2x^{2}+5x-3=9 \\ \\ 2x^{2}+5x-12=0 \\ D = 25 + 4*2*12 = 25 +96 = 121 \\ \\ x_{1} = \dfrac{-5+11}{4} = \dfrac{3}{2} \ ; \ x_{2} = -4$
Проверка: 
$x = \dfrac{3}{2}: \ \ log_{_{3}}(2 *1.5-1) + log_{_{3}}(1.5+3) = 2 \ \rightarrow \ 2 = 2 \\ \\ x = -4: \ \ log_{_{3}}(2*(-4)-1) + log_{_{3}}(-4+3) = 2 \ \rightarrow \ Reshenia\ net$
Ответ: x = 1.5

#2
$log_{_{0,2}}x\leq -3 \\ log_{_{5^{(-1)}}}x\leq -3 \\ -log_{_{5}}x\leq -3 \ \ /*(-1) \\ x \geq 5^{3} \\ x \geq 125 \\ x\in [125;+\infty)$
Ответ: x∈[125;+∞)