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

$log_2^2(3x+1)-3log_{frac{1}{2}}frac{4}{3x+1}=Big (frac{2}{7}Big )^{log_{frac{2}{7}}1,5+log_{frac{2}{7}}4}\\ODZ:; ; left { {{3x+1>0} atop {frac{4}{3x+1}>0}} right. ; ; to ; ; 3x+1>0; ,; x>-frac{1}{3}\\log_2^2(3x+1)+3log_2frac{4}{3x+1}=Big (frac{2}{7}Big )^{log_{frac{2}{7}}_1,5}}cdot Big (frac{2}{7}Big )^{log_{frac{2}{7}}4}\\log_2^2(3x+1)+3(log_24-log_2(3x+1))=1,5cdot 4\\log_2^2(3x+1)+6-3log_2(3x+1)=6\\log_2(3x+1)cdot (log_2(3x+1)-3)=0\\a); ; log_2(3x+1)=0; ,; 3x+1=1; ,; x=0\\b); ; log_2(3x+1)=3; ,; ; 3x+1=8; ,; x=frac{7}{3}=2frac{1}{3}\\Otvet:; ; x=0; ,; x=2frac{1}{3}; .$