Statistical Physics: very large and very small numbers

[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?

 

[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?

 

[PhysicsGirl90]

They have a different base. So if we wanted to compare them we they would both have to have the same base.

 

[BruceW]

exactly. So what can you do to get them both to have the same base?

 

[PhysicsGirl90]

Thank you for your comment Bruce...i figured it out...convert (1/44) to a power of 10 and the rest follows easily.

 

[PhysicsGirl90]

(1/44)^(10^5) = (10^log(1/44))^(10^5)...Thanks for helping me think it through.

 

[BruceW]

yeah, no worries. Glad to have helped :)