I almost said the same thing, but when I looked back I realized that "find the number of digits" means, clearly, "find the exact (integer) number of digits in the number". If the number had thousands of digits, a rough estimate would be appropriate, but the 2^10=10^3 approximation makes an "exact" answer easy for reasonable numbers, since you are, after all, rounding the log up to the nearest integer. Integer answers can easily be exact. And, of course, there is no need to have an exact value for the power, in order to find the number of digits.
I agree with one small caveat. The 2^10 approximates 10^3 has a sizable relative error, in excess of 2%. Without doing a formal analysis on propagation of error, I'd hesitate before claiming an exact answer on any computation with a relative error of that magnitude even with the extra knowledge that the answer must be an integer.
SK's idea of bounds avoids the question of propagation of error.
\(\displaystyle 2^3 < 10 \text { and } 2^{10} > 10^3.\)
\(\displaystyle \therefore 2^{230} = 2^{20} * 2^{210} = (2^10)^2 * (2^10)^21 > (10^3)^2 * (10^3)^{21} \implies\)
\(\displaystyle 2^{230} > 10^69.\)
Therefore there are at least 70 meaningful digits in the decimal representation of 2^230.
\(\displaystyle 2^{230} = 2^2 * 2^{228} = 4 * (2^3)^{71} < 4 * 10^{71}.\)
Therefore there are at most 72 meaningful digits in the decimal representation of 2^230.
You can perhaps do better without a calculator, computer, or table of logarithms, but you can get the answer of 71 digits plus or minus 1 based on just arithmetic.
But as you say, you can do much better with logarithms. Except in very unusual cases where the log is very close to being an integer, the order of magnitude can be calculated exactly.
\(\displaystyle y = 2^{230} \implies log_{10}(y) = 230 * log_{10}(2) \approx 69.2 << 70.\)
Therefore, the exact answer is 70 digits.
I do wonder, however, what is trying to be taught with these problems. Order of magnitude is important, but the two low order digits of 2017^2017 seem to be inconsequential.