How do you solve this?

2log(x) = log(4x + 3)     note that we can write this as

log ( x² ) = log(4x + 3)     and equating expressions, we have

x² =  4x + 3                     rearranging, we have

x² - 4x - 3 = 0               no factoring is available here, so using the onsite solver, we have

$${{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{4}}{\mathtt{\,\times\,}}{\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{3}} = {\mathtt{0}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{2}}{\mathtt{\,-\,}}{\sqrt{{\mathtt{7}}}}\\
{\mathtt{x}} = {\sqrt{{\mathtt{7}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{0.645\: \!751\: \!311\: \!064\: \!590\: \!6}}\\
{\mathtt{x}} = {\mathtt{4.645\: \!751\: \!311\: \!064\: \!590\: \!6}}\\
\end{array} \right\}$$

Note something here...if we write this equation as 2log(x) = log(4x + 3), then we are explicitly assuming that x > 0  and only the positive solution is correct. However, if we write the equation as  log(x²) = log(4x + 3), then both solutions are correct  !!!