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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.

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    $\begingroup$ Well, $2^{1500}$ in binary is 1 followed by 1500 zeroes. $\endgroup$ – gt6989b Nov 4 '13 at 14:36
  • $\begingroup$ I don't think there is a such method, in fact $x=2^{1500}$ is just infinite even for the computer. $\endgroup$ – Martial Nov 4 '13 at 14:39
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    $\begingroup$ Not really, my computer says it's 35074662110434038747627587960280857993524015880330828824075798024790963850563322203657080886584969261653150406795437517399294548941469959754171038918004700847889956485329097264486802711583462946536682184340138629451355458264946342525383619389314960644665052551751442335509249173361130355796109709885580674313954210217657847432626760733004753275317192133674703563372783297041993227052663333668509952000175053355529058880434182538386715523683713208549376 $\endgroup$ – Dan Shved Nov 4 '13 at 14:41
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    $\begingroup$ @Christoph I don't know, I think it is doable in a couple of days, maybe a week, if you're smart about it. $\endgroup$ – Dan Shved Nov 4 '13 at 14:44
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    $\begingroup$ My first remark would be that $\log_{10} 2 \approx 3/10$ so your number has about $1500 \cdot 3/10 = 450$ digits. For more than that you have to do actual work... $\endgroup$ – TMM Nov 4 '13 at 14:50
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$\log_{10} 2 \simeq 0.301030$, so $\log_{10} 2^{1500} = 1500 \log_{10} 2 \simeq 451.545$, and from this, we can know that $2^{1500}$ is a 452-digit number in base 10.

Since the decimal part of $\log_{10} 2^{1500}$ is $0.545...$, $\log_{10} 3 \simeq 0.4771$ and $\log_{10} 4 \simeq 0.301030*2 = 0.602060$, we can know that $2^{1500} \simeq 3 \times 10^{451}$. ($\log_{10} (x \times 10^{451}) = \log_{10} x + \log_{10} 10^{451} = \log_{10} x + 451$) Indeed, it is $3.507... \times 10^{451}$. If a log table is provided, you memorized some entries, or you calculated log values of various constants (probably using Taylor expansion of $\ln x$?), then you can get more digits.

If you want to get accurate values of $2^{1500}$, instead of calculating $2^1, 2^2, \cdots, 2^{1499}$ sequentially, you can use the fact that $2^{2k} = (2^{k})^2$ to minimize your work. For example, if one knows that $2^{25} = 33554432$ (it's quite easy to memorize!), one can calculate $2^{50} = 33554432^2 = 1125899906842624$, without calculating $2^{26} \cdots 2^{49}$.

(Of course, you still can calculate it by doubling previous values; multiplying big numbers is more complex and easier to have a mistake than doubling a number.)

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If one only need a quick and very rough estimate, one should find a number close to what you need to estimate and yet you know how to handle. You then expand the result from the number you know.

For the case $2^{1500}$, one can use the fact $2^{10} = 1024$ is very close to a power of $10$ and $\log 2 \sim 0.7$. We have $$\begin{align} 2^{1500} = & (2^{10})^{150} = (1.024 \times 10^3)^{150}\\ \sim & e^{0.024\times 150} \times 10^{450} = e^{3.6} \times 10^{450} = e^{5\times 0.7 + 0.1} \times 10^{450}\\ \sim & 2^5 e^{0.1} \times 10^{450} = 3.52 \times 10^{451} \end{align}$$ Compare with the correct answer $\sim 3.50746 \times 10^{451}$, this estimate is accurate up to around $1\%$. Not bad for a calculation you can do it in your head.

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It depends how much energy you have. In the old days a few people calculated things of this size (452 decimal digits). Repeated squaring can get you $2^{1024}$ in just $10$ multiplications, then you pick the seven you need and multiply them and you are there. Of course, some of these multiplications are a lot of work.

If you are happy with approximations, going through logs makes it pretty easy. It helps to have some log tables available.

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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).

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According to pari/gp, the precise value of this is

35074662110434038747627587960280857993524015880330828824075798024790963850563322203657080886584969261653150406795437517399294548941469959754171038918004700847889956485329097264486802711583462946536682184340138629451355458264946342525383619389314960644665052551751442335509249173361130355796109709885580674313954210217657847432626760733004753275317192133674703563372783297041993227052663333668509952000175053355529058880434182538386715523683713208549376

You might use logarithms to calculate the first few digits (if you have $\log_{10} 2$ sufficiently accurately) and you can use congruence facts to find the last few digits, but for the full value, you are going to have to do a lot of work using pencil and paper!

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  • $\begingroup$ After posing the question, I realized that it was a silly one since how can I expect to solve this monster by pen and paper. But is there a way of approximating it like $3.5*10^55$ or something like that using log? $\endgroup$ – Eames Cobb Nov 4 '13 at 14:44
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    $\begingroup$ I started to explain, but JiminP's typing is faster ... $\endgroup$ – Old John Nov 4 '13 at 14:47

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