0\\ 2x>-1\\ x>-\frac{1}{2}\\\\ \log_3\sqrt{2x+1}=1\\ 3^1=\sqrt{2x+1}\\ 3=\sqrt{2x+1}\ |^2\\ 9=2x+1\\ 2x=8\\ x=4\\ " alt="\\\log_3\sqrt{2x+1}=1\\ 2x+1>0\\ 2x>-1\\ x>-\frac{1}{2}\\\\ \log_3\sqrt{2x+1}=1\\ 3^1=\sqrt{2x+1}\\ 3=\sqrt{2x+1}\ |^2\\ 9=2x+1\\ 2x=8\\ x=4\\ " align="absmiddle" class="latex-formula">