The largest n-bit number is \(2^n - 1\). The number of decimal digits in it is about \(\log_{10}(2^n) = n\log_{10}2\), rounded up.

So \(d = \lceil n\log_{10}2 \rceil\), which means \(d \ge n\log_{10}2\) (usually strictly greater). For example, n = 10 gives 1023, with d = 4 > 3.01.

Answer: (D) \(d > n\log_{10}2\)