Consider a positive integer number $2^{{10}^n}$ where $n\geq 5$.
Question: How to prove the length of the number $2^{{10}^n}$ is equal to $30103{\underbrace{00\cdots0}_{n-5}}$.
Example: Consider $n=6$ then we have $$ 2^{{10}^6}=990065622929589825069792361\cdots 71236104888403162747109376 $$
It can be checked that the length of $2^{{10}^6}$ is $301030$.
Try:
Consider the following approximation $$ 2^{10}=1024 \approx 10^3 \quad \Rightarrow \quad 2^{10^n} \approx 10^{3\times10^{n-1}} $$
It is easy to check that the length of $10^{3\times10^{n-1}}$ is equal to $3{\underbrace{00\cdots0}_{n-1}}$.
Thanks for any suggestion
The number of digits in a positive integer $n$ is $\lfloor\log_{10}n\rfloor+1.$
$0.30102<\log_{10} 2<0.30103$
so $30102<\log_{10} 2^{100000} < 30103;\quad$ i.e., $2^{100000}\approx{10^{30103}}.$
Therefore, the number of digits in $2^{100000}$ is $30103$.
However, actually $\log_{10} 2= 0.30102999566...,$ so the number of digits in $2^{10^9}$ is $301029996,$
not $301030000$ as you thought.