Question

$\frac{1}{2} log_{2}( x^{2}+x+2)- log_{2}(3x-2)=1$

Answer 1

$\sqrt{x^2+x+2} =2*(3x-2) \\x\ \textgreater \ \frac{2}{3} \\ x^{2} +x+2=12x-8 \\ x^{2} -11x+10=0 \\ x_{1}=1; x_{2}=10$