Logarithm rules i.e. $-1/2 \cdot \log_2(2/9)$ to $2/9\cdot \log_2(2)$? I'm struggling to understand the flow of calculation as shown in the picture below.

It would be really nice if someone can explain how does one reach from step one to step two and which Logarithm rules were applied here to reach the second step from the first one.
Thank you! :)

I'll give you the steps, but you should probably familiarize yourself with the (very simple) rules: https://www.chilimath.com/lessons/advanced-algebra/logarithm-rules/.
\begin{align}
&1 - \frac{-\frac{1}{3}\log_2\left(\frac{1}{3}\right) - \frac{1}{2}\log_2\left(\frac{2}{9}\right)}{-\sum\limits_{s = 1}^9 \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
&\textrm{Denominator: You're just adding the same thing 9 times}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(\left(\frac{1}{3}\right)^{-1}\right) + \frac{1}{2}\log_2\left(\left(\frac{2}{9}\right)^{-1}\right)}{-9 \cdot \frac{1}{9}\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{-\log_2\left(\frac{1}{9}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(\left(\frac{1}{9}\right)^{-1}\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(\frac{9}{2}\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\left(\log_2\left(9\right) - \log_2\left(2\right)\right)}{\log_2\left(9\right)}\\\\
=\ &1 - \frac{\frac{1}{3}\log_2\left(3\right) + \frac{1}{2}\log_2\left(9\right) - \frac{1}{2}}{\log_2\left(9\right)}\\\\
\end{align}
For this to be equal to the second step, you'd need $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} = \frac{2}{9}\log_2\left(2\right) = \frac{2}{9}$ to be true. It's not: $\frac{1}{2}\log_2\left(9\right) - \frac{1}{2} \approx 1.1$ and $\frac{2}{9} \approx 0.2$.
