How do I Approximate $\log{2}\approx 0.693$ without using the Maclaurin series? How  do I approximate the value $\log{2}\approx 0.693$ without using the Maclaurin series?
The book gives the hint: consider $f(x)=e^x-e^{-x}-2x$.

If $y=\ln(x)$, then $$y'=\frac{1}{x}\quad\text{and}\quad y'=e^{-y}$$ and you can numerically solve either differential equation to $x=2$, using initial condition $y(1)=0$. 
With the first differential equation, using the Runge-Kutta method with only two steps:
$$
\begin{array}{rrr|rrrr}
n&x_n&y_n&k_{1,n}&k_{2,n}&k_{3,n}&k_{4,n}\\
0&1&0\\
&&&\frac{1}{1}=1&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{4}}=\frac{4}{5}&\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\\
1&\frac32&0+\frac{\frac{1}{2}}{6}\left(1+2\cdot\frac{4}{5}+2\cdot\frac{4}{5}+\frac{2}{3}\right)=\frac{73}{180}\\
&&&\frac{1}{\frac{3}{2}}=\frac{2}{3}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{4}}=\frac{4}{7}&\frac{1}{\frac{3}{2}+\frac{1}{2}}=\frac{1}{2}\\
2&2&\frac{73}{180}+\frac{\frac{1}{2}}{6}\left(\frac{2}{3}+2\cdot\frac{4}{7}+2\cdot\frac{4}{7}+\frac{1}{2}\right)=\frac{1747}{2520}
\end{array}
$$
And $\frac{1747}{2520}=0.693\ldots$. For more accuracy, you have to start over with a smaller step size (and hence, more steps).
