\[\begin{array}{l}
 {\log _{2x + 1}}({x^3} + 3{x^2} - 6x) \cdot {\log _x}(2x + 1) = 3 \\ 
 {\log _{2x + 1}}({x^3} + 3{x^2} - 6x) \cdot \frac{1}{{{{\log }_{2x + 1}}x}} = 3 \\ 
 \frac{{{{\log }_{2x + 1}}({x^3} + 3{x^2} - 6x)}}{{{{\log }_{2x + 1}}x}} = 3 \\ 
 {\log _x}({x^3} + 3{x^2} - 6x) = 3 \\ 
 \end{array}\]