Question

Решить логарифмическое неравенство (подробно)

Answer 1

решение на фотографии

Answer 2

$log _{2}(dfrac{1}{x} -1) + log _{2} ( dfrac{1}{x} + 1) le log _{2}(27x-1) \ \ ODZ: left{ begin{gathered} dfrac{1}{x} -1 > 0 \ dfrac{1}{x}+1>0 \ 27x - 1 > 0 \ x ne 0 end{gathered} right. ; left{ begin{gathered} x > 0 \ x < 1 \ x > dfrac{1}{27} \ x > -1 end{gathered} right. (1)$

$x in ( dfrac{1}{27} ; 1)$

$log _{2}(dfrac{1}{x} -1) + log _{2} ( dfrac{1}{x} + 1) le log _{2}(27x-1) \ \ log _{2}((dfrac{1}{x} -1)(dfrac{1}{x} + 1)) le log _{2}(27x -1) \ \ log _{2}(dfrac{1}{x^{2}} -1) le log _{2}(27x-1) \ \ dfrac{1}{x^{2}} - 1 le 27x -1 \ \ dfrac{1}{x^{2}} - 1 - 27x +1 le 0 \ \ dfrac{1}{x^{2}} - 27x le 0 \ \ dfrac{1-27x^{3}}{x^{2}} le 0 / * x^{2}, T.K. x^{2} > 0 \ \ 1 - 27x^{3} le 0 \ \ 27x^{3} ge 1 \ \ x ge dfrac{1}{3}$

С учётом ОДЗ (2):

$x in [dfrac{1}{3} ; 1)$

Ответ: x ∈ [1/3 ; 1)