Problema Solution
Andrew measures the amount of a very unstable substance to be 100 moles. The half life of this substance is three days (after three days, half is gone).
#1 Write an exponential function that models this situation where y is the amount of substance and x is time in days.
#2 Use your equation to complete the chart.
Days Amount
-3(3 days prior) ?
0 ?
3 ?
6 ?
9 ?
12 ?
#3 Use your data from the chart to graph your equation.
#3 Calculate the expected amount of substance if Andrew had taken his measurement 9 days earlier. Show your calculations and consider the trend of your graph.
Answer provided by our tutors
#1
Let A(t) denotes the amount of material in time t
A(t) = A(0)*e^(kt)
A(0) = 100
the half life is 3 days thus
A(3) = (1/2) A(0)
A(3) = A(0)*e^(3k)
A(0)*e^(3k) = (1/2) A(0) divide both sides by A(0)
e^(3k) = 1/2
3k = ln (1/2)
k = (1/3) ln 0.5
A(t) = 100*e^((1/3)*t*ln(0.5))
#2
-3(3 days prior) A(-3) = 100*e^((1/3)*(-3)*ln(0.5)) = 100 e^(- ln 0.5) =100 (e^ln 0.5)^(-1) = 100*(0.5)^(-1) = 200 mol
0 ? A(0) = 100*e^((1/3)*0*ln(0.5)) = 100 mol
3 ? A(3) = 100*e^((1/3)*3*ln(0.5)) = 100*e^(ln 0.5)) = 100*0.5 = 50 mol
6 ? A(6) = 100*e^((1/3)*6*ln(0.5)) = 100*e^(2 ln 0.5)) = 100*(0.5)^2 = 25 mol
9 ? A(9) = 100*e^((1/3)*9*ln(0.5)) = 100*e^(3 ln 0.5)) = 100*(0.5)^3 = 12.5 mol
12 ? A(12) = 100*e^((1/3)*12*ln(0.5)) = 100*e^(4 ln 0.5)) = 100*(0.5)^4 = 6.25 mol