[help me] Let m and n denote the number of digits in $2^{2007}$ and $5^{2007}$ when expressed in base 10. What is the sum m + n?
1. To determine the number of digits in \(2^{2007}\) and \(5^{2007}\), we need to use the formula for the number of digits of a number \(n\) in base 10, which is given by:
   \[
   d(n) = \lfloor \log_{10} n \rfloor + 1
   \]
2. First, we calculate the number of digits in \(2^{2007}\):
   \[
   d(2^{2007}) = \lfloor \log_{10} (2^{2007}) \rfloor + 1 = \lfloor 2007 \log_{10} 2 \rfloor + 1
   \]
   Using the approximation \(\log_{10} 2 \approx 0.3010\):
   \[
   2007 \log_{10} 2 \approx 2007 \times 0.3010 = 603.107
   \]
   Therefore:
   \[
   d(2^{2007}) = \lfloor 603.107 \rfloor + 1 = 603 + 1 = 604
   \]
3. Next, we calculate the number of digits in \(5^{2007}\):
   \[
   d(5^{2007}) = \lfloor \log_{10} (5^{2007}) \rfloor + 1 = \lfloor 2007 \log_{10} 5 \rfloor + 1
   \]
   Using the approximation \(\log_{10} 5 \approx 0.6990\):
   \[
   2007 \log_{10} 5 \approx 2007 \times 0.6990 = 1403.893
   \]
   Therefore:
   \[
   d(5^{2007}) = \lfloor 1403.893 \rfloor + 1 = 1403 + 1 = 1404
   \]
4. To find the sum \(m + n\), where \(m\) is the number of digits in \(2^{2007}\) and \(n\) is the number of digits in \(5^{2007}\), we add the results from steps 2 and 3:
   \[
   m + n = 604 + 1404 = 2008
   \]

The final answer is \(\boxed{2008}\).