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This is pretty straightforward, but I'd like to study, how find out, how many times do you need to double previous result of calculation to get some sum, for example: $10^{82}$

  1. $1\times 2 = 2$
  2. $2\times 2 = 4$
  3. $4\times 2 = 8$
  4. $8\times 2 = 16$

n. $x\times 2 \geq 10^{82}$

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up vote 5 down vote accepted

The $n$th step is $2^n$. If you want $2^n\geq A$, then you want $$n = \log_2(2^n) \geq \log_2(A) = \frac{\ln(A)}{\ln(2)} = \frac{\log_{10}A}{\log_{10}(2)}.$$

So the first $n$ at which $2^n\geq A$ will be the least positive integer greater than or equal to $\log_2(A)$, which is denoted $$\left\lceil \log_2A \right\rceil.$$

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...and, in particular, $A = 10^{82}$ implies $\log_2 A = \log_{10} A / \log_{10} 2 \approx 82 / 0.30103 \approx 272.4$, and thus $\lceil \log_2 A \rceil = 273$. – Ilmari Karonen Jun 28 '12 at 21:00
Added with the last comment, this was what I was looking for. Thanks. – MarkokraM Jun 29 '12 at 5:34

For good ballpark estimates, use the fact that $2^{10}=1024\approx 1000$. So $10$ doublings is about the same as multiplying by $1000$. Since $10^{82}=10\times 1000^{27}$, we can see that $270$ doublings will get us past $10^{81}$. There is some slack, and another $3$ doublings get us over.

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While I saw this also very interesting and working approach, I accepted other answer because it gave more precise and general solution for my purpose. Thanks. – MarkokraM Jun 29 '12 at 5:32

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