We have three responses for you
Use the fact that 2^{10} is approximately 10^{3}. (In fact, 2^{10} = 1024.) That makes 2^{330} in the neighborhood of 10^{99}. Thus, the answer to your question is that 10^{100} requires about 333 binary digits. If you want to be more accurate you will have to work harder.
Chris
Using a calculator:
2^{3.321}= 9.993569...
2^{3.322} = 10.0005...
So, 10^{100}> (2^{3.321})^{100} = 2^{332.1}
And, 10^{100}< (2^{3.322})^{100} = 2^{332.2}
So, in the binary expansion of of 10^{100}, there will be a digit (0 or
1) for each of the following places: 2^{0}, 2^{1}, 2^{2}, ..., 2^{332}
Hence, there are 333 digits
Paul
The number of digits of a number is revealed when you look at the logarithm of the number using a base equal to the number of possible digits. For example, 243 (in normal decimal notation  base 10) has three digits. When we take the logarithm in base 10, we see log_{10 }243 2.39. We ignore the decimal portion and add one to the whole number and get 3 digits. Let's try this using binary.
243 in base ten is 11110011 in base two (binary). This is frequently written
243_{10 } = 11110011_{2.}
Now take the logarithm in base two: log_{2 }243 7.92. Again, we drop the .92 and then add one to the seven and our method gives the correct 8 digits.
So your question can be answered if you add one to the whole number portion of
log_{2 }10^{100 }. Remember to use the shortcuts that logarithms permit with exponents.
Hope this helps,
Stephen La Rocque.
