Can someone tell me to get e to the other side of the equation you multiply it by ln, as well as the other side. What happens to '-mt' in this instance?
((Ti-T0)/(T1-T0))=e^-mt
Thanks in advance
I don't quite understand your post, but the natural log isn't a number that you multiply. You can take the natural log of an expression and when you have an equation you can take the ln of both sides and maintain equality, but it isn't a multiplication process.
As for the right hand side:
$\displaystyle \ln(e^{-mt})=(-mt)\ln(e)=(-mt)(1)=-mt$
This is applying one of the 3 common log rules to manipulate terms from outside the log with inside.