Calculate $2^{1500}$ Is there a way of approximating $2^{1500}$ using pen and paper? I just took $2^{1500}$ as an example but it can be $2^{150000}$ or $12^{95}$. I am just looking for pen and paper method.
By approximate, I mean a solution like $3.5*10^{21}$ or something along those lines.
I was thinking of doing the following: $2^{1500} =x$ then take the $log$ of both sides but I don't know if that makes sense.

Well, if I was forced to do it by hand, I would do it this way. I would first express $1500$ as a sum of powers of $2$, like this: $ 1500 = 1024 + 256 + 128 + 64 + 16 + 8 + 4$. Then I would find, one after another, these powers of $2$: $2^1, 2^2, 2^4, 2^8, \ldots, 2^{1024}$. Each next power of two is calculated as the square of the previous one, using decimal multiplication as taught in school. Then I would "simply" multiply $2^{1024} \cdot 2^{256} \cdot 2^{128} \cdot 2^{64} \cdot 2^{16} \cdot 2^8 \cdot 2^4$.
This is just a rough sketch. In reality I would spend the first couple of hours refining the method itself to minimize the number of digit multiplications. Maybe it would be wiser to not calculate those powers of 2 and multiply them afterwards, but rather calculate numbers in this sequence:
$$
2^4 \to
2^{4+1} \to
2^{8 + 2} \to
2^{8 + 2 + 1} \to
2^{16 + 4 + 2} \to
2^{16 + 4 + 2 + 1} \to
2^{32 + 8 + 4 + 2} \to
2^{64 + 16 + 8 + 4} \to
2^{64 + 16 + 8 + 4 + 1} \to
2^{128 + 32 + 16 + 8 + 2} \to
2^{128 + 32 + 16 + 8 + 2 + 1} \to
2^{256 + 64 + 32 + 16 + 4 + 2} \to
2^{256 + 64 + 32 + 16 + 4 + 2 + 1} \to
2^{512 + 128 + 64 + 32 + 8 + 4 + 2} \to
2^{1024 + 256 + 128 + 64 + 16 + 8 + 4} = 2^{1500}.
$$
If you do it like this, at each step you either multiply the current number by $2$, or square it. You don't multiply two distinct large numbers with each other.
This isn't conceptually different from the first approach. But this way the amount of long multiplications is a bit less, so it can save you a day or two.
Again, I would compare these two approaches and calculate more precisely the amount of digit-by-digit multiplications involved in each method. Then I would go with the one that turns out to be shorter (my intuition tells me it's the second one).
