text: # Homework Help: Statistical Physics: very large and very small numbers
  
  1. Jun 10, 2012
  
  ### PhysicsGirl90
  
  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. Jun 10, 2012
  
  ### BruceW
  
  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?
  
  3. Jun 10, 2012
  
  ### PhysicsGirl90
  
  They have a different base. So if we wanted to compare them we they would both have to have the same base.
  
  4. Jun 10, 2012
  
  ### BruceW
  
  exactly. So what can you do to get them both to have the same base?
  
  5. Jun 10, 2012
  
  ### PhysicsGirl90
  
  Thank you for your comment Bruce...i figured it out...convert (1/44) to a power of 10 and the rest follows easily.
  
  6. Jun 10, 2012
  
  ### PhysicsGirl90
  
  (1/44)^(10^5) = (10^log(1/44))^(10^5)...Thanks for helping me think it through.
  
  7. Jun 10, 2012
  
  ### BruceW
  
  yeah, no worries. Glad to have helped :)