Question

log^2 2 (3x+1) - 3log 1/2 (4/3x+1)=(2/7)^(log 2/7 1,5 + log 2/7 4)

Answer 1

$log_2^2(3x+1)-3log_{\frac{1}{2}}\frac{4}{3x+1}=\Big (\frac{2}{7}\Big )^{log_{\frac{2}{7}}1,5+log_{\frac{2}{7}}4}\\\\ODZ:\; \; \left \{ {{3x+1>0} \atop {\frac{4}{3x+1}>0}} \right. \; \; \to \; \; 3x+1>0\; ,\; x>-\frac{1}{3}\\\\log_2^2(3x+1)+3log_2\frac{4}{3x+1}=\Big (\frac{2}{7}\Big )^{log_{\frac{2}{7}}_1,5}}\cdot \Big (\frac{2}{7}\Big )^{log_{\frac{2}{7}}4}\\\\log_2^2(3x+1)+3(log_24-log_2(3x+1))=1,5\cdot 4\\\\log_2^2(3x+1)+6-3log_2(3x+1)=6\\\\log_2(3x+1)\cdot (log_2(3x+1)-3)=0\\\\a)\; \; log_2(3x+1)=0\; ,\; 3x+1=1\; ,\; x=0\\\\b)\; \; log_2(3x+1)=3\; ,\; \; 3x+1=8\; ,\; x=\frac{7}{3}=2\frac{1}{3}\\\\Otvet:\; \; x=0\; ,\; x=2\frac{1}{3}\; .$