If we know that $\ln(9.6)\approx 2, 26$, and that $\ln(0.3)\approx -1.20$, is it possible to approximate $\ln(0.5)$?
Since $\ln(9.6) = \ln(\frac{96}{10}) = \ln(96)-\ln(10)\approx 2.26$, and $\ln(0.3) = \ln(\frac{3}{10}) = \ln(3)-\ln(10)\approx -1.20$, we see that $\ln(32) = \ln(96)-\ln(3)\approx 3.46$. Therefore, $\ln(2)\approx 0.692$, and $\ln(0.5) = \ln(\frac{1}{2}) = \ln(1)-\ln(2)\approx -0.692$.
You don't need $\ln(64)$. Rather $\ln(32)=\ln(2^5)=5\ln(2)$. Continue from there.
Subtle ... and significant
After the above was posted, the OP proceeded to edit the question so the answer is now revealed as $-0.692$.
In fact the actual logarithm to three significant figures is $-0.693$. Even though standard rules for significant figures were obeyed by rendering tge result to just three digits, the third significant digit fails to match reality.
The subtraction operation is not ill-conditioned here because $\ln(9.6)$ and $\ln(0.3)$ have opposing signs. But the output has a larger first sgn8ficant digit than the input values ($6$ versus $1$ and $2$). Therefore the third significant digit in the output is guaranteed only if the percentage error is finer than that guaranteed by the given inputs. When following significant-digit takes in scientific calculations, beware of this effect. You may have to allow for uncertainty in your last significant digit.
