нужно срочно please)))). Created by maikochanturia. algebra-ru

нужно срочно please))))...

maikochanturia @maikochanturia

May 2022 1 10 Report

нужно срочно please))))

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ОДЗ : x > 0

$log_{3}\frac{1}{x}=log_{3}\frac{1}{27}-log_{3}81\\\\log_{3} \frac{1}{x}=log_{3}3^{-3}-log_{3}3^{4} \\\\log_{3}\frac{1}{x}=-3-4\\\\log_{3}\frac{1}{x}=-7\\\\\frac{1}{x}=3^{-7}\\\\\frac{1}{x}=\frac{1}{2187} \\\\x=2187$

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maikochanturia May 2022 | 0 Ответы

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maikochanturia May 2022 | 0 Ответы

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