I have been working through the following exercise (chapter $0$, number $17$, in Weissman's An Illustrated Theory of Numbers):
How many bits does the binary expansion of $10^{500}$ have?
I know the number of binary digits is $\lfloor 500 \cdot \log_2 10 \rfloor + 1$, so I can get the answer with a calculator.
Could I know how many bits the binary expansion of $10^{500}$ has, without a calculator or a logarithmic table? Does it help that this exercise considers the number of binary digits of a power of $10$?
It's easy to observe that $10^{500} = 5^{500} \cdot 2^{500}$, so its binary expansion ends with $500$ zeros. Then, I used the following factorization of $5^{500} - 1$: \begin{equation*} \begin{split} 5^{500} - 1 &= (5^{250} + 1) (5^{250} - 1) = (5^{250} + 1) (5^{125} + 1) (5^{125} - 1) \\ &= (5^{250} + 1) (5^{125} + 1) (5 - 1) (5^{124} + \ldots + 5 + 1). \end{split} \end{equation*} Therefore, the largest power of $2$ dividing $5^{500} - 1$ is $2^4$ and the binary expansion of $5^{500}$ ends with $10001$. But I can't get any further.
Maybe you know that $\log_{10}2 \approx 0.30103$. We have $\log_2{10}=\frac 1{\log_{10}2}\approx \frac 1{0.30103}$. Doing the inversion is just a long division, which is quite possible without a calculator.
To compute $\log_2 (10)$ you can start with $\log_2(10)=3+\log_2(1.25)$. Take successive square roots of $2$ by hand and find the first one less than $1.25$. It will be $2^{1/4}$, so $\log_2(10)=3.25+\log_2(\frac {1.25}{2^{1/4}})$ You can keep going until you get the accuracy you need. You need it within about $0.002$. It will be some work, but you can get there. Tables are wonderful things.