Подобно все и понятно.[tex] log_{2}( log_{3}( log_{2}(2x + 1)) = 3[/tex]. Created by BpexLoader. algebra-ru

Пишем ОДЗ:Значит хє(0.5;+∞)Решаем по типу:Это и есть ответ.

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Подобно все и понятно.
[tex] log_{2}( log_{3}( log_{2}(2x + 1)) = 3[/tex]

Answers & Comments

QDominus

Verified answer

$log_{2}( log_{3}( log_{2}(2x + 1) ) ) = 3$

Пишем ОДЗ:

$log_{3}( log_{2}(2x + 1) )>0\\log_{2}(2x + 1)>1\\2x+1>2\\2x>1\\x>0.5$

Значит хє(0.5;+∞)

Решаем по типу:

$log_{a}(b) = c \\ b = {a}^{c}$

$log_{2}( log_{3}( log_{2}(2x + 1) ) ) = 3 \\ log_{3}( log_{2}(2x + 1) ) ) = {2}^{3} \\ log_{2}(2x + 1) = {3}^{ {2}^{3} } \\ 2x + 1 = {2}^{ {3}^{8} } \\ 2x = {2}^{ {3}^{8} } - 1 \\ x = \frac{ {2}^{ {3}^{8} } - 1 }{2}$

Это и есть ответ.

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