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I know some basic laws of logs, such as $$c \cdot \log(x) = \log(x^c)$$ $$\log(xy) = \log(x) + \log(y)$$ $$\log(\frac{x}{y}) = \log(x) - \log(y)$$

But I just can't see how $$\frac{1}{2} \cdot (1 - p^2-q^2) \cdot \log_2(1-p^2-q^2)$$ turns into $$pq \cdot \log_2(2pq)$$

EDIT: This is from a probability paper on coin flipping, so that $q = 1-p$.

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    It doesn't. Take $p = q = 1$ as a counterexample. Are there any restrictions on $p$ and $q$? – 2017-01-28
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    @tilper, yes sorry. It's from a paper on probability (getting fair flips from biased coins), so $q = 1-p$ – 2017-01-28
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    `q=1-p` Hint: that gives $1-p^2-q^2=2pq$. – 2017-01-28
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    @dxiv, embarrassing...thank you. Still find the downvote a little rough (not directed at you). – 2017-01-28

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$$p+q=1 \implies (p+q)^2=1 \implies p^2+q^2+2pq=1 \implies 1-p^2-q^2=2pq$$ Thus $$\frac{1}{2} \cdot (1 - p^2-q^2) \cdot \log_2(1-p^2-q^2)=pq\log_{2}(2pq)$$

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    Thank you for writing it up as an answer, I just got done figuring it out based on @dxiv's comment. I can't believe I spent 45 minutes trying every algebraic manipulation on this thing without even remembering that $q$ and $p$ were defined in relation to each other... – 2017-01-28
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    @jeremyradcliff That's all right. :) We all make mistakes from time to time. – 2017-01-28
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    You're very nice, which reminds me I should be nicer to myself :) – 2017-01-28