Interesting right?

We note that:

\(\begin{array} {} \log_{10} 0.5 & = \log \dfrac{5}{10} & = 0.69897 - 1 & = - \color{blue}{0}.30103 & \Rightarrow \color{blue}{0} \text{ zero} \\ \log 0.05 & = \log \dfrac{5}{100} & = 0.69897 - 2 & = - \color{blue}{1}.30103 & \Rightarrow \color{blue}{1} \text{ zero} \\ \log 0.005 & = \log \dfrac{5}{1000} & = 0.69897 - 3 & = - \color{blue}{2}.30103 & \Rightarrow \color{blue}{2} \text{ zeros} \\ ... & ... & ... & ... & ... \\ \log \left(\dfrac{1}{2} \right)^{1000} & = 1000 \log \dfrac{5}{10} & = 1000(0.69897-1) & = -\color{blue}{301}.03 & \Rightarrow \boxed{\color{blue}{301}} \text{ zeros} \end{array} \)

It can be written as (5/10)^1000 which means after the decimal point there are 1000 numbers after the decimal point. 5^1000 has ceiling function of 1000log5 digits=699. So of the 1000 numbers after decimal point, 699 are non-zero and remaining are 0's', so there are 1000-699=301 zeroes between the decimal point and the first non zero digit.

Given $log_{10} 5 = .69897$ we should rearrange this so it relates to 1/2. 1/2 = .5.

$log_{10} (5 /10) = log_{10} (.5) = (log_{10} 5) - log_{10}10 = .69897 - 1 = -.30103$

Plugging back into the original we have: $\frac{1}{10^.30103} = \frac{1}{2}$ .

Raising to the 1000^th: $\frac{1}{10^{301.03}} = \frac{1}{2^{1000}}$ .

So now we must realize that $\frac{1}{10^n}$ produces $\ceil{n-1}$ zeros after the decimal.

$\frac{1}{10^{301.03}}$ produces $\ceil{301.03 - 1} = \boxed{301}$ zeros after the decimal.

First, we take log in base 10 to find the number of zeroes. $\log (\frac{1}{2})^{1000}$ $=-1000 \log 2$

We can find $\log 2$ using the hint given by noting that $\log 2 = \log \frac {2*5}{5} = \log 10 - \log 5 \approx 0.30103$

$-1000 \log 2 \approx -301.03$ so there are $\boxed {301}$ 0's .