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I'm struggling to understand the flow of calculation as shown in the picture below.

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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! :)

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    $\begingroup$ While the other substitutions are clear, the replacement you emphazized in the title does not make much sense to me. I mean clearly $-\frac{1}{2} \log_2 \left(\frac{2}{9}\right) \neq \frac{2}{9}\log_2 2 = \frac{2]{9}$. $\endgroup$ – dfnu Jan 7 at 20:46
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    $\begingroup$ The substitution isn’t correct. $\endgroup$ – KM101 Jan 7 at 20:48
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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 $\frac{2}{9} \approx 0.2$.

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