Exponential + Logarithm Equation

$\large 3^x \cdot 2^{\frac{3x}{x+2}}=2\cdot 3$ $\large \implies \dfrac{2^{\frac{3x}{x+2}}}{2}=\dfrac{ 3}{3^x}$ $\large \implies 2^{\frac{2(x-1)}{x+1}}=3^{1-x}$ Taking $\ln$ on both sides: $\dfrac{\cancel{(x-1)}}{x+2}\ln 2^2=-\cancel{(x-1)}\ln 3$ Here we can clearly see $\color{#3D99F6}{\boxed{x=1}}$ is one solution. $\large \implies x+2=-\dfrac{\ln 4}{\ln 3}=-log_3 4$ $\large\implies \color{#3D99F6}{ \boxed{x=-2-2\log_{3} 2}}$

Therefore sum of solution= $1+(-2-2\log_{3} 2)=-1-2\log_3 2$ $\therefore 1+2+3+2=\huge \color{#D61F06}{\boxed 8}$