# reducing fractions with logs in the numerator

I was reading a tutorial introduction to Information Theory and it presented a formula for determining 'average surprise' of 100 coin flips with 50 heads and 50 tails, as follows:

$${[50 * log(1/0.5)] + [50 * log(1/0.5)] \over 100}$$

The text goes on to say that the above expression reduces to:

$$[0.5 * log(1/0.5)] + [0.5 * log(1/0.5)]$$

My question is, why did the log(1/0.5) not reduce to a smaller number in the same way that 50 reduces to 0.5 ? I intuitively expect that it should reduce to:

$$(0.5 * {log(1/0.5) \over 100}) + (0.5 * {log(1/0.5) \over 100})$$

More generally, how are logs in the numerator simplified? Thank you.

• Your question is unclear. The whole expression reduces further, so just perform the substitutions. What really are you asking? – David G. Stork Aug 31 '18 at 19:04
• Paul D, 1/0.5=...? – user376343 Aug 31 '18 at 19:14
• Attempted to clarify my question. – Paul D Aug 31 '18 at 19:25

First, you can do $$\frac {50a+50b}{100}=\frac {50(a+b)}{100}=0.5(a+b)$$ Then $1/0.5=2$ so I don't know why the book did not substitute $$\log \frac 1{0.5}=\log 2$$ to get $$\log 2$$ as the final answer.
• I updated the wording of my question to be hopefully clearer. I get what you are illustrating above, but I still don't see why ${50 + log(a) \over 100 }$ does not reduce to $0.5 + { log(a) \over 100}$ – Paul D Aug 31 '18 at 19:35
• They have $50$ times $\log 2$, not $50+\log 2$. Your reduction in the comment would be correct but that is not the problem you posted. – Ross Millikan Aug 31 '18 at 19:41