Given that \[\log_{10} 2 = 0.3010 \; \text{ (to }4\text{ d.p.)}\] and that \[10^{0.2} < 2\] it is possible to deduce that

\(2^{100}\) begins in a \(1\) and is \(30\) digits long;

\(2^{100}\) begins in a \(2\) and is \(30\) digits long;

\(2^{100}\) begins in a \(1\) and is \(31\) digits long;

\(2^{100}\) begins in a \(2\) and is \(31\) digits long.

From the first given property, we have that \[\log_{10} 2^{100} = 100\log_{10} 2 = 30.10\dotsc\] so \[2^{100} = 10^{30.10 \dotsc} = 10^{0.10\dotsc} \times 10^{30}\] and so \(2^{100}\) is \(31\) digits long, with the first digit being the first digit of \(10^{0.10\dotsc}\).

Now we use the second given property: \[1 < 10^{0.10\dotsc} < 10^{0.2} < 2.\] The first digit is less than \(2\) and so must be a \(1\), and the answer is (c).