How one makes a nice approximation using Taylor series Approximations of expressions like $2^{0.6}$ , or $3^{0.7}$ . For firrst one i know binomial expansion can help with x very less we need but is it better to do taylor expansion around √2 than binomially expansion if we just need upto three decimal places ? ( I know the log values needed for taylor expansion so basically is it better to do taylor expansion around √2 or some better value , or binomial is best , if we meed upto 3 decimal places)

Write $2^{0.6}$ as $\exp(0.6 \ln 2)$ where $\exp(x) = e^x$, where you know $\ln 2$ to three decimal places.
Then since $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + O(x^5)$, the maximum Lagrange remainder is the next term, thus $\frac{(0.6 \ln 2)^n}{n!} > 0.001$ for any power of $n$. Some trial and error gives $n  > 4$, so we need the $x^5$ term as well:
$$2^{0.6} \approx1 + 0.6 \ln 2 + \frac{1}{2!}(0.6 \ln 2)^2 + \frac{1}{3!}(0.6 \ln 2)^3 + \frac{1}{4!}(0.6 \ln 2)^4 + \frac{1}{5!}(0.6 \ln 2)^5$$
and the approximation with $\ln 2 \approx 0.693$ gives $1.51558$, whereas the actual value of $2^{0.6}$ is around $1.51572$, an absolute error of around $1.4 \times 10^{-4}$, and a relative error of $9.3 \times 10^{-5}$.
