Solve $\log_3(3^x+3)\log_3(3^{x+1}+19)=12$

Question

My approach is to extract those $x$ from those logarithm even though I dun know what should I do next, but unfortunately I still can't extract it. Here is how I do it: \begin{equation} \log_3(3^x+3)\log_3(3^{x+1}+19)=12\\ \Rightarrow 3^{\log_3(3^x+3)\log_3(3^{x+1}+19)}=3^12 \end{equation} From the above equation I could extract only one from those two namely $(3^x+3)$ or $(3^{x+1}+19)$ using $a^{\log_a{b}}=b$. Please help

Answer 1

If we ignore those pesky constants in the logs this becomes $x(x+1)=12$ with the obvious solution $x=3$. The constants will increase the left side, so we must have $x \lt 3$. If we try $x=2$ the left side becomes $\log_3(12)\log_3(46)\lt 3 \cdot 4 =12$ so $x \gt 2$. If there isn't an integral solution, there probably isn't a pretty one, so I fed it to Alpha, getting $x \approx 2.8394482194330276320$