I am confused. What you show as the "equation ... given in the book"wouldbe correct if "beta" and "delta" were constants- but "beta" is not constant in this problem. A complete solution to this problem, as given, would involve solving a "separable differential equation".

Personally, for "half life" problems I prefer "If the half life is T then N(t)= N_0*(1/2)^{t/T}". That is, t/T counts the number of "half lives" and the amount halves for each of those.( http://www.wolframalpha.com/input/?i=2*e^%28%281%2F%28t%2B1%29+-+1%2F10%29*t%29+%3D+1*e^%28%282%2F%28t%2B2%29+-+1%2F10%29*t%29 )

P_1 = P_2 should be when one population surpasses another, since P_2 is growing faster than P_1 this should be the critical point.

However, I have set this up incorrectly.

3.

Let k = half life. Half life is decreasing so constant k is negative.

Let t = elapsed time.

Let N_0 = the number of atoms at t=0

Let N(t) = the number of atoms total.

Given equations:

1/2 * N_0 = N_0 * e^(-k*t)

N(t) = N_0 * e^(-k*t)

That can be converted to e^{-kt} by using "x= e^ln(x)" as (1/2)^{t/T}= e^ln((1/2)^{t/T}= e^(t/T)ln(1/2) so that "-k= ln(1/2)/T" (since 1/2< 1, ln(1/2) is negative).

Your error is in thinking that kMy work:

N(4) = 6000

6000 = N_0 * e^(-8*4)

N_0 = 6000/(e^(-8*4))

This is not equal to 6000 * sqrt(2). My issue appears to be in my setup.isthe half life. It is not that easy. Using that set up, knowing that the half life is 8, you know that for given N(0),

N(8)= N(0)e^{-k(8)}= (1/2)N_0 (i.e. half as much)

Dividing both sides by N_0, e^{-8k}= 1/2. Taking the ln of both sides, -8k= ln(1/2)= -0.6931 (to four decimal places) so k= -0.6931/8= -.0866.

Solve 6000= N_0e^{-.0866(4)}

Or, using my formula, with half life T= 8, 6000= N_0(1/2)^{4/8}= N_0(1/2)^(1/2) so that N_0= 6000/(1/2)^(1/2)

Again, your error is in thinking that the "k" in the formula5.

Let S_0 be the starting amount in the account.

Let k be the interest rate.

Let t be the time elapsed.

Let S(t) be the current amount in the account.

S(t) = s_0*e^(k*t) (given equation)

S(t)_1 = s_0*e^(.03*2 + .03*t) (We're starting with two years elapsed)

S(t)_2 = s_0*e^(.05*t)

Setting S(t)_1 = S(t)_2,

e^(.03*2 + .03*t) = e^(.05*t), t = 3.isthe interest rate. With interest rate "r", an initial amount "S" will have increased to S+ rS= (1+ r)S in one year. So we would have to have Se^{k(1)}= Se^k= (1+ r)S so that e^k= 1+ r. From that k= ln(1+ r), not r. If r= .03, k= ln(1.03)= 0.02956 while if r= .05, k= ln(1.05)= 0.04879.