Converting $5\log$ reduction to a percentage

I see a lot of terms bounded about like a 5log reduction of bacteria. So I presume it's $5\log_{10}$. Now how do I convert that into a percentage?

Would it be:

\begin{align} 5\log_{10}10=&\displaystyle \left(1-\dfrac{x}{100}\right)100\\ \Rightarrow x\%=&\displaystyle 100\left(1-\dfrac{1}{100}5\log_{10}10\right) \end{align}?

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If the $\log_{10}$ is reduced by $5$, the quantity is reduced by a factor $10^{-5}$. This is $0.001\%$ of the original quantity, or a $99.999\%$ reduction
@user1134241: I'm not sure what you are looking for. Maybe $\log_{10}(new)-\log_{10}(old)=-5 \implies \frac{new}{old}=10^{-5}$ –  Ross Millikan Mar 13 '13 at 15:52
Looks good, thank you. I was after something like a conversion formula between % and log so eg: What log reduction (x) is 55%. $x=-log(1-\dfrac{55}{100})\quad \Rightarrow\quad x\simeq0.347$? –  HCAI Mar 13 '13 at 15:57
@user1134241: You have understood it correctly. You have said $55\%$ is a reduction of $55\%$ or to $45\%$ of the original amount. Various people use percentages in different ways. –  Ross Millikan Mar 13 '13 at 16:02