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I realize this was asked before, but I need more precision. For example a good approximation for $2^{256}$ = $1.2$ x $10^{77}$. I've followed various methods, but the best I can do is get $1.02$ x $10^{77}$. So here is one of the methods I tried. I need to use the approximation $10^3 ≈ 2^{10}$

$2^{256}$ = $2^{10 * 25}$ * $2^{6}$
$10^{3*25}$ = $10^{75}$
$64$ x $10^{75}$ or $6.4$ x $10^{76}$

off to the side
$25.6$ x $2.4$ = 61.44
61.44 / 100 = 0.6144

6.4 * 0.6144 = 3.93216

6.4 + 3.93216 = 10.33216

$2^{256}$ = $1.03$ x $10^{77}$

Is there a way to get this more precise using the $10^3$ approximation technique?

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  • $\begingroup$ in actual fact it's 1.157 ... $\endgroup$ – Roddy MacPhee Apr 30 at 10:51
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You cannot really use a better approximation than that because $2^{10}, 2^{20}, 2^{30}$ successively diverge farther from $10^{3n}$, you could tell.

However, it is worth noting that $2^{40}=1.099.511.627.775$, which is extremely close to $1.1\times10^{12}$ .

You can then take the sixth power, getting $2^{240}\approx1.77\cdot10^{72}$.

Then, $2^{250}\approx1.77\cdot10^{75}$.

Multiplying by $64$, we get $1.13\cdot 10^{77}$.

Not much better, but slightly. It also involves more computation with the $1.1$ part.

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  • $\begingroup$ I guess my efforts where in vain then, thanks. $\endgroup$ – Travis Apr 28 at 19:30
  • $\begingroup$ yeah i am sorry but like even at 2^50 no clean approximations $\endgroup$ – Saketh Malyala Apr 28 at 19:32

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