Put \(u=e^x>0\), with \(u\ne3\). Multiplying by \(u(u-3)\), \[u^2+2=4u(u-3)\Rightarrow3u^2-12u-2=0.\] Thus \[u=\frac{6\pm\sqrt{42}}3.\] The minus root is negative and cannot equal \(e^x\). The positive root differs from 3, so \[x=\ln\left(\frac{6+\sqrt{42}}3\right)=1.426.\]