$ You write it in it's normal form: $
\frac{\log{(\frac{x+2}{2})}}{\log{\sqrt{a}}} = \frac{\log{(2x)}}{\log{a}}
$ Using the fact that: $ \log{a^x} = x \log{a}
\frac{2\log{(\frac{x+2}{2})}}{\log{a}}=\frac{\log{(2x)}}{\log{a}}
\frac{(x+2)^2}{4}=2x $ Taking note that $ x > 0
$ Solve...