Hey bottlerockets, I just saw ur question I think this is the answer:
I=Ioe^ev/kt
u then divide both sides by Io to get:
I/Io=e^ev/kt
u then (Ln) both sides to get:
Ln(I/Io)=ev/kt
In logarithms if ur havin Ln(x/y), then itz the same as Lnx-Lny, so:
LnI-LnIo=ev/kt
To make the equation in y=mx+c form:
LnI=ev/kt+LnIo
y=LnI
m=e/kt
c=LnIo
Hope this helps, sorry if the symbols are not clear!