Question

ЛОГАРИФМЫ! НАЙТИ ПРОИЗВЕДЕНИЕ КОРНЕЙ УРАВНЕНИЯ
log(9x)3 * log(9)25 = log(3)5 * log(x)3 * log(3x)3

Answer 1

$log_{9x}3\cdot log_925=log_35\cdot log_{x}3\cdot log_{3x}3\; ;\; \; ODZ:\; x\ \textgreater \ 0,x\ne 1\\\\\frac{1}{log_39x}\cdot log_{3^2}5^2=log_35\cdot \frac{1}{log_3x}\cdot \frac{1}{log_3{3x}}\\\\\frac{1}{log_39+log_3x}\cdot 2\cdot \frac{1}{2}\cdot log_35=log_35\cdot \frac{1}{log_3x\cdot (log_33+log_3x)}\\\\\frac{1}{2+log_3x}=\frac{1}{log_3x+log_3^2x}\\\\log_3x+log_3^2x=2+log_3x\\\\log^2_3x=2\; \; \Rightarrow \\\\log_3x=-\sqrt2\; \; \; ili \; \; \; log_3x=+\sqrt2\\\\x=3^{-\sqrt2}\; \; \; ili \; \; \; x=3^{\sqrt2}$

$3^{-\sqrt2}*3^{\sqrt2}=3^0=1$