I am asking because I believe the following question:
Evaluate $66^e$
was a part of a non-calculator exam that permitted decimal approximations. (Only 3 decimal places were needed)
I do not believe that the question is possible because $e$ is irrational and it seems impossible to calculate a power with irrational exponent. However, I cannot be sure.
On
In order to accomplish any approximation work one has to establish an approximate value possessing the right order of magnitude, from which the actual approximation can be started. In the case at hand we start from $2^{16}\approx 64^e$.
Write $66=64\cdot\bigl(1+{1\over32}\bigr)$, so that $66^e=64^e\left(1+{1\over32}\right)^e$.
Use the approximation $e\doteq2.7183$, hence $6e\doteq 16.31$. It follows that $$64^e=2^{6e}\doteq2^{16}\cdot2^{0.31}=65\,536\cdot\left(1+{1\over3}\right)^{0.31}\cdot\left(1-{1\over3}\right)^{-0.31}\ .$$ In this way we obtain $$66^e\doteq65\,536\cdot\left(1+{1\over3}\right)^{0.31}\cdot\left(1-{1\over3}\right)^{-0.31}\cdot\left(1+{1\over32}\right)^e\ .$$ The three powers on the right hand side can be computed with the binomial series. Taking terms up to $(1/3)^4$ and up to $(1/32)^2$ we obtain the value $$65\,536\cdot1.09318\cdot1.13328\cdot1.08723=88\,273.6\ .$$ The true value is $88\,314.7$.
You can approximate:
$$e\approx2.718281828$$
I recommend memorizing this if you can't have a button that gives you the value of $e$ since it has the nice $18281828$ part, super easy to remember IMO.
Thus, we have
$$66^e\approx66^{2.7}=66^{27/10}$$
And then I imagine you have your ways to approximate this (?)
Indeed, this is how I'd imagine you'd handle most irrational exponents.
A second method is to take the natural log:
$$66^e=e^{e\ln(66)}$$
The $\ln(66)$ can easily be handled with Taylor's theorem, as can $e^x$.