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Statistical Physics: very large and very small numbers

  1. Jun 10, 2012 #1
    Working on statistical physics i came across this expression:

    p = (1/44)^(10^5) = 10^(-164345)

    However TI-83 calculator is unable to verify it (gives answer 0). Can someone tell me how to get from (1/44)^(10^5) to 10^(-164345) analytically?
     
  2. jcsd
  3. Jun 10, 2012 #2

    BruceW

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    You are still going to need to use a calculator at some point (their answer of 10^(-164345) is not exact, it has been rounded off). But yes, there is a way to find the answer which the calculator can cope with.

    To start with, what is the main difference between the expressions (1/44)^(10^5) and 10^(-164345) ? Like if you wanted to compare the two numbers, what would be the first thing you would do?
     
  4. Jun 10, 2012 #3
    They have a different base. So if we wanted to compare them we they would both have to have the same base.
     
  5. Jun 10, 2012 #4

    BruceW

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    exactly. So what can you do to get them both to have the same base?
     
  6. Jun 10, 2012 #5
    Thank you for your comment Bruce...i figured it out...convert (1/44) to a power of 10 and the rest follows easily.
     
  7. Jun 10, 2012 #6
    (1/44)^(10^5) = (10^log(1/44))^(10^5)...Thanks for helping me think it through.
     
  8. Jun 10, 2012 #7

    BruceW

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    yeah, no worries. Glad to have helped :)
     
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