RSA square and multiply
After reading the comments on your original post, I thought I'd give you an example for exponentiating by $e$. Take for example the case $e=2^{16}+1$. Then, to calculate $x^e$ we use $16$ squaring operations and one multiply. This is because 
$$x^{2^{16}+1}=x \cdot x^{2^{16}} = x\cdot x^{\underbrace{2\cdot 2 \cdots 2}} =x\cdot ((x^2)^2)\cdots)^ 2$$
(where the brace contains sixteen $2$'s).

So, we use repeated squaring to calculate $x^{2^{16}}$, then having done so can simply multiply by $x$ once to find $x^e$.

If we think about the binary expansion of $e$, this techinque can be generalised to give [relatively] efficient exponentiation to any power. For example, how to we take an element to the power $e=41$? Well, first writing $e$ in terms of powers of two (ie in binary), we see that $e=32+8+1=2^5+2^3+2^0$ and thus $e=101001$. So, to evaluate this we simply initialise our answer as $1$ and set $y=x$ ($y$ will store $x^{2^i}$ as $i$ increases), then work from right to left along the binary expansion of $e$. If the digit is $1$, then we multiply by $y$. Then, we square $y$ and move on to the next digit.

The reason $e$ is chosen to be a value like $2^{16}+1$ or $3=2^1+1$ are that they each only require one multiplication. Looking back at my rather poor explanation above, we see that we have to square $x$ for each 'place' in the binary expansion of $e$, and we have to multiply our running value once for each $1$ in the binary expansion of $e$. So, if $e$ is $n$ bits long (ie $n=\lfloor\log_2(e)\rfloor+1$) we require $n-1$ squaring oparations and at most $n$ multiplications (or $n-1$ if we start our multiplication with the first value power of $x$ encountered rather than initialising with $1$ as in my example).

As mentioned by other comments, it is hard to know how many multiplications/squarings will be required to decrypt, since to do so requires knowing the binary expansion of $d$. That said, we can be pretty confident that $d$ is very unlikely to have anywhere near as nicer expansion as $e$ did with these choices, suggesting decrypting will require rather more work than encryption.