Given \( P=P_{0} e^{-k t} \) When \( t=36, P=12 \) and when \( t=11 \) then \( P=30 \) Determine the values of the constants \( k \) and \( P_{0} \).

We are given the equation   P = P₀ e^(–k t) with two data points: when t = 36 the population P = 12, and when t = 11 the population P = 30. Step 1. Write the equations for the two data points. For t = 36:   P₀ e^(–36k) = 12       (1) For t = 11:   P₀ e^(–11k) = 30       (2) Step 2. Divide equation (1) by equation (2) to eliminate P₀.   (P₀ e^(–36k))/(P₀ e^(–11k)) = 12/30 Simplify the left-hand side:   e^(–36k + 11k) = e^(–25k) So we have:   e^(–25k) = 12/30 = 2/5 = 0.4 Step 3. Solve for k by taking the natural logarithm of both sides.   ln(e^(–25k)) = ln(0.4)   –25k = ln(0.4) Thus, the value of k is:   k = –(ln 0.4)/25 Since ln(0.4) is negative, k is positive. An alternative form is:   k = (ln(1/0.4))/25 = (ln 2.5)/25 Step 4. Find P₀ using one of the original equations. We use equation (2):   P₀ e^(–11k) = 30   P₀ = 30 e^(11k) Substitute the expression for k:   P₀ = 30 e^(11(ln 2.5)/25) This is the exact expression for P₀. You could also express it as:   P₀ = 30 · 2.5^(11/25) Step 5. (Optional) Approximate the values. First, approximate k:   ln 2.5 ≈ 0.91629   k ≈ 0.91629/25 ≈ 0.03665 Now, approximate P₀:   P₀ = 30 e^(11·0.03665) = 30 e^(0.40315)   e^(0.40315) ≈ 1.4963   P₀ ≈ 30 × 1.4963 ≈ 44.89 Final Answer:   k = (ln 2.5)/25       and  P₀ = 30 e^(11(ln 2.5)/25)   (Approximately, k ≈ 0.03665 and P₀ ≈ 44.89)