Question

$log_{2}3+log_{3}(x^2+3x+5)=2$

Answer 1

$log_{2}3+log_{3}(x^2+3x+5)=2\\ x^2+3x+5>0\\ D=9-4*1*5<0\\ frac{1}{log_{3}2}+log_{3}(x^2+3x+5)=2\\ log_{3}(x^2+3x+5)=2-frac{1}{log_{3}2}\\ x^2+3x+5=3^{2-frac{1}{log_{3}2}}\\ x^2+3x+5=frac{9}{3^{frac{1}{log_{3}2}}}\\ x^2+3x+5 - frac{9}{3^{frac{1}{log_{3}2}}}=0\\ D=9-4*1*(5-frac{9}{3^{frac{1}{log_{3}2}}}<0$
 нет решений