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What is the smallest integer integer $m$ such that $2^m > 10^{21}$?

I determined that since $2^{50} > 5^{20}$, we see that $2^{71} > 10^{20}$. The answer is $m = 70$, but I haven't figured out a way to show it is minimal. Also I need to do this without a calculator.

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    $\begingroup$ Logarithms make short work of such problems. This problem in particular amounts to finding $\lceil \log_2 10^{21} \rceil$. It is of course possible to evaluate this by pencil and paper, so I suggest you clarify what "without a calculator" requires. $\endgroup$ – hardmath Jul 30 '16 at 3:59
  • $\begingroup$ @hardmath I mean just with pencil and paper. How would we solve it without a calculator? $\endgroup$ – Puzzled417 Jul 30 '16 at 4:01
  • $\begingroup$ @Puzzled417 Anything doable in a calculator is, given enough time, possible to do on the paper, hence the request for clarification. $\endgroup$ – Git Gud Jul 30 '16 at 4:02
  • $\begingroup$ One method for extracting logarithms "by pencil and paper" requires taking square roots. Are you familiar with pencil and paper methods for this? $\endgroup$ – hardmath Jul 30 '16 at 4:06
  • $\begingroup$ @hardmath No, can you explain? $\endgroup$ – Puzzled417 Jul 30 '16 at 4:07
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$2^{10}=1.024\times 10^3$

So $2^{70}=(1.024)^{7}\times 10^{21}$

Now we just have to prove $1.024^{7}\leq 2$

By the binomial theorem

$1.1^7=1+.7+.21+.035+.0035+.00021+.000007+.0000001=1.9487171$

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    $\begingroup$ The essence here is familiar to anyone who works with computers: $2^{10}$ is just a touch larger than $10^3$, so $2^{70}$ is a bit larger than $10^{21}$, and some calculation is needed to ensure that it isn't so much larger that $2^{69}$ will also work (it's quite far from being that large). $\endgroup$ – Erick Wong Jul 30 '16 at 4:16
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So $2^{10}=1024=10^3\cdot1.024$ so $2^{70}=10^{21}\cdot 1.024^7>10^{21}$

Now $\sqrt 2 > 1.4$ so $2^{1/8}>1+0.4^3 = 1.064$ so $1.024^7<2 \implies 2^{69}<10^{21}$

The same line of reasoning would lead me to infer that $2^{159}<10^{48}<2^{160}$

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Do you know that $\log_{10}2\approx 0.30103$? Then you are looking for $0.30103m \gt 21$ and you can just divide.

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    $\begingroup$ No calculator he said. $\endgroup$ – Zack Ni Jul 30 '16 at 4:01
  • $\begingroup$ @ZackNi: A number of people know that value. I didn't have to look it up. Many classes that prohibit calculators expect you to memorize a bunch of constants. $\endgroup$ – Ross Millikan Jul 30 '16 at 4:32
  • $\begingroup$ Modulo physics, every mathematics constant derives from math formula. For example, $\pi$ derives from $\int_0^1 \sqrt{1-x^2} \times 4$ or $e$ is a series of factorial numbers. Even some really simple constant like $\sqrt{2}\approx 1.414$, which comes from binary search and multiplication. If you rottenly memorize those mathematics constant, the math axiom will collapse and you will be immortal. To put it in a nutshell, it should reveal what kind of method you use to approximate it rather than remembering stiffly. $\endgroup$ – Zack Ni Jul 30 '16 at 4:49
  • $\begingroup$ @ZackNi: I disagree strongly. Knowing $\pi \approx 3.1416 \approx \sqrt {10}$ makes many approximate computations easier. The more constants you know, and the more simple things like $(1+x)^n \approx 1+xn$ for $nx \ll 1$, the more you can do. I don't see why you should have to derive the value of $\pi$ or $\sqrt 2$ when you need one. It is also useful to know series that can derive them, true. A bigger toolbox can often be an advantage. I also disagree with no calculator exams, but I don't play the game so don't make the rules. $\endgroup$ – Ross Millikan Jul 30 '16 at 5:01
  • $\begingroup$ Yes you can disagree strongly but I cannot see anything beneficial from using a calculator in exam. It will make a lot of student leak their step in their reasoning. So bad. I agree the more you know, the more you solve but the more constant you know, the less derivative of the constant you will try. When it comes to not use a calculator (stated in the problem given by OP), you should think you are ignorant to all constant and give the derivative of the constant. $\endgroup$ – Zack Ni Jul 30 '16 at 5:15

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