<s>[INST] Is there an algorithm to find the number of digits in 2^n, where n is a positive integer? Is there an algorithm to find the number of digits in 2^2030 ?
$2^1=2$
$2^2=4$
$2^3=8$
$2^4=16$
$2^5=32$
$2^6=64$
$2^7=128$
...
$2^{10}=1024$
 [/INST] The number of digits in $2^n$ is $1+floor(n \log_{10} 2)$.
Using $\log_{10} 2 \approx 0.30103$ gives you 612 digits.
Here is an explanation:
A number $x$ has $k$ digits in base 10 iff $10^{k-1} \le x < 10^k$. Taking logs, we get $k-1 \le  \log_{10} x < k$, which means that $floor(\log_{10} x)=k-1$. So, $k=1+floor(\log_{10} x)$.
An answer to the question in the title needs high-precision approximations of $\log_{10} 2$.
</s>