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