How do I Approximate $\log{2}\approx 0.693$ without using the Maclaurin series?

If you start from $\ln(2)=\int_1^2\frac{1}{x}\,dx$, then Simpson's Rule with $n=4$ gets you there fast:

$$\ln(2)\approx\frac{1}{12}\left(1+4\left(\frac45\right)+2\left(\frac23\right)+4\left(\frac47\right)+\frac12\right)\approx0.693\ldots$$

Surely not what the book's author had in mind, but still instructive:

On a calculator with a $\sqrt{\phantom2}$ button you could approximate $\log c$ by repeatedly hitting $\sqrt{\phantom2}$, getting a sequence of numbers each about twice as close to $1$ as the one before. After $n$ iterations we reach $c^{2^{-n}}$, which for large $n$ is about $1 + 2^{-n} \log c$. So subtract $1$ and multiply by $2^n$ to approximate $\log c$. On a 10-digit calculator you can get 5-digit accuracy (confirmed by comparison with the output of the "$\ln$" button...) by iterating enough times to get within about $10^{-5}$ of $1$, which is to say $n \cong 16.6 + \log_2 \ln c$ [with $16.6$ arising as $\log_2 (10^5)$].

For $c=2$ we get $n \cong 16$, and indeed calculation of $$ 2^{16} (2^{2^{-16}} - 1) = 2^{16} \left(\sqrt{\sqrt{\sqrt{\cdots\sqrt 2}}} - 1 \right) $$ yields a number between $0.6931$ and $0.6932$.