Without using Wolfram Alpha, evaluate #log_10(y) = -138.265#? The answer is not #0#.

It's easy to do it in Wolfram Alpha, but how could I simplify it so that it's able to be evaluated on a TI-83 calculator? There's a limit to how large or small the displayed numbers can be.

This is useful, for example, when you evaluate #DeltaS = k_BlnOmega#, the microstates equation for entropy.

1 Answer
Aug 25, 2016

Hm, you know what, it would easier to just evaluate part of it on a calculator, and do the rest of it using factors of #10^10#.

#10^(-138.265) = y#

#=> 10^(-38.265 - 100) = y#

#=> 10^(-38.265)/(10^(100)) = y#

Since #10^(-38.265) = 5.433xx10^(-39)#, #color(blue)(y = 5.433xx10^(-139))#.