0 ≤ t ≤ ln(3)
When t = 0
[tex]s(0) = e^0 = 1[/tex]
when t = ln(3)
[tex]s(ln(3)) = e^{ln3} = 3[/tex]
We have two coordinates: (0, 1) and (ln(3), 3)
The rate of change = The difference in y-coordinates ÷ the difference in x-coordinates
The rate of changes = [3 - 1] ÷ [ln(3) - 0] = 2/ln(3)
Answer: 2/ln(3)