Question

Логарифм
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Answer 1

$ODZ: \frac{2}{1-x}\ \textgreater \ 0\\\\1-x\ \textgreater \ 0\\\\x\ \textless \ 1$

$\displaystyle log_{11}log_3log_2 \frac{2}{1-x}=0\\\\log_3log_2 \frac{2}{1-x}=11^0\\\\log_3log_2 \frac{2}{1-x}=1\\\\log_2 \frac{2}{1-x}=3\\\\ \frac{2}{1-x}=2^3\\\\ 1-x= \frac{2}{8}\\\\x=1- \frac{2}{8}\\\\x= \frac{3}{4}$