exponentiation
I'd like to answer this question without using a calculator.
My first instinct is to find a pattern in the value of $2^n$:
$2^1 \equiv 2$, $2^2 \equiv 4$, $2^3 \equiv 8$, $2^4 \equiv 6$, $2^5 \equiv 2,…$
So units digit form a set of $4$ digits: $2,4,8,6,2,4,8,6…$
Now I'm stuck, how do I know which digit to pick from the set above
This is what I would do to calculate $e^x$ with perfect 8-digit accuracy.
Take $\lfloor x\rfloor$ and $b = x - \lfloor x\rfloor$, the whole and fractional parts respectively. ($e^x = e^{\lfloor x \rfloor + (x - \lfloor x\rfloor)} = e^{\lfloor x \rfloor} e^{x - \lfloor x \rfloor}$)
Use the $(((b/4 + 1)b/3 + 1)b/2 + 1)b + 1$ pattern to calculate $e^{x - \lfloor x\rfloor}$. (If the fractional part of the exponent is .4 or less, only terms up to $b/7$ are needed for 8-digit accuracy. Worst-case scenario ($b \rightarrow 1$) terms up to $b/10$ are needed.) This method is easy to implement on a simple calculator, especially ones with a memory slot to quickly reinsert $b$.
When that is finished, multiply by $e$ ($\approx 2.71828183$ by memorization) and press the equals button $\lfloor x\rfloor$ times to repeat multiply. The result is $e^x$.
I've analyzed the number of terms required for full accuracy. Here is the chart (click to view full image):
Basically, if terms up to $b/t$ are needed to fully calculate $e^x$ up to 8-digit precision (7 digits after the decimal point), accounting for rounding (the $\frac{1}{2}$ term), the relation between $x$ and $t$ is given by $\sqrt[t]{\frac{1}{2}10^{-7}t!} = x$.
$$2.2^{3.3}\gtrless3.3^{2.2}$$ $$(2.2^3)^{1.1}\gtrless (3.3^2)^{1.1}$$ $$2.2^3\gtrless 3.3^2$$ $$(2\cdot 1.1)^3\gtrless (3\cdot 1.1)^2$$ $$2^3\cdot 1.1\gtrless 3^2$$ $$8\cdot 1.1\gtrless 9$$ $$8.8<9.$$
So, $$2.2^{3.3}<3.3^{2.2}.$$
Can you find now which number is larger: $2.25^{3.375}$ or $3.375^{2.25}$ ;) ?
Best Answer
As you can see, it repeats with period $4$. The units digit of $2^k$ thus depends on $k \pmod 4$. Can you find $102 \pmod 4$?