I = Alog10(t) + B
I equals 8 when t is 3
I equals 50 when t is 8
t can't go past 200
A and B are constants
Find A and B when I reaches 100
8=Alog3+B
50=Alog8+B
subtract the first from the second
42=a(log8-log3)
42=a(log(8.3))
solve for A
then
hmmm. wondering if ou meant when t=100...