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Large numbers

  1. Nov 24, 2012 #1
    which is greater as n gets large, [itex] f(n) = 2^{2^{2^n}} [/itex] or [itex] g(n) = 100^{100^n} [/itex]

    instinctively I'd go with f(n) but I have no idea of actually showing that f(n) would indeed get larger, obviously sticking in values of n doesn't particularly work. A method I was thinking was to show many digits would f(n) be when n=100, and same with g(n), I attempted this but to no avail,

    any hints would be appreciated

    thanks.
     
  2. jcsd
  3. Nov 24, 2012 #2

    Dick

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    Take the log of both functions. What do you think now? If you're not sure take the log again.
     
    Last edited: Nov 24, 2012
  4. Nov 26, 2012 #3
    [tex]ln(2^{2^{2^n}}) = 2^{2^n}ln(2) [/tex]
    [tex]ln(2^{2^n}ln(2)) = ln(2^{2^n}) + ln(ln(2)) = 2^nln(2) + ln(ln(2)) [/tex]

    [tex]ln(100^{100^n}) = 100^nln100 [/tex]
    [tex]ln(100^nln(100)) = ln(100^n) + ln(ln(100)) = nln(100) + ln(ln(100)) [/tex]

    from this I'm sure that 2^n will indeed grow much larger as it's an exponential however I don't particularly know how to conclude.

    Also, is there anyway I can do it by looking at the digits?
     
  5. Nov 26, 2012 #4

    Dick

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    Sure. Now it's pretty obvious 2^n will be much larger than n for n large. If you want to be formal about it you could use l'Hopital's theorem on the ratio of the ln(ln)'s. But I don't think you have to. And this is really the same as counting digits. To find the number of digits you would look at log base 10 of each expression. When you take the second log you are basically looking at the log of the number of digits.
     
  6. Nov 26, 2012 #5
    Thanks for your help
     
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