Precisamos trocar as bases dos logs, portanto:
[tex]Log_3\ (a-2)\ -\ Log_\ \frac{1}{3}\ (a-2) =2[/tex]
[tex]\frac{Log_\frac{1}{3}\ (a-2)}{Log_\frac{1}{3}\ 3} - Log_\frac{1}{3}\ (a-2) = 2[/tex]
[tex]Log_\frac{1}{3}\ 3 =x \\\\\ \frac{1}{3}^x = 3\\\ 3^-^x=3^1\\\\ x=-1[/tex]
[tex]-Log_\frac{1}{3}\ (a-2) - Log_\frac{1}{3}\ (a-2) = 2\\\\ -2\ Log\frac{1}{3}\ (a-2) =2\\\\\ Log\ \frac{1}{3}\ (a-2) =-1\\\\ \frac{1}{3}^-^1 = (a-2)\\\\\ 3 = a-2\\\\ \boxed{\therefore\ a=5}[/tex]