How many digits are in the integer representation of 2 to the 30th power?
Since I didn't really know any 'expert' way to approach this, I just started out by listing the powers of 2, like 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024 etc. and the same sequence looking only at the digits would be 1, 1, 1, 2, 2, 2, 3, 3, 3, 4 etc. At first it'd seem like the pattern is 3 numbers for each digit (if you understand what I mean), but from 1024 onwards, the pattern breaks, with there being 4 four-digit numbers, 3 five-digit numbers, 3 six-digit numbers and et cetera.
So I couldn't find a pattern, and now I'm stuck, but I think there might actually be an easier way to solve this, or maybe a formula.
Best Answer
For the first few hundred powers of two, it's roughly true that $10^{3k} \approx 2^{10k}$
Specifically, if $k=3$, that gives that $2^{30}$ is a little over a billion, which means it has $\boxed{10}$ digits.
A general formula for the number of digits of any power of $2$ can be found using the value of the base-$10$ logarithm of $2$. The number of digits in $2^k$ is
$$1 + \left\lfloor k\log_{10}(2)\right\rfloor$$