The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula $x = De^{-0.2t}$ where $x$ is the amount of the antibiotic in the bloodstream in milligrams, $D$ is the dose given in milligrams and $t$ is the time in hours after the antibiotic has been given - Edexcel - A-Level Maths Pure - Question 2 - 2015 - Paper 3

First, we need to calculate the amount of antibiotic in the bloodstream just after the first dose (after 5 hours):

$x_1 = 15 e^{-0.2 imes 5}$

Calculating:

$-0.2 imes 5 = -1.0$ $x_1 = 15 e^{-1} \approx 15 \times 0.367879 = 5.018$

Next, for the second dose given after 5 hours, we compute:

$x_2 = 15 e^{-0.2 imes (5 + 2)} = 15 e^{-1.4}$

Calculating:

$-0.2 imes 7 = -1.4$ $x_2 \approx 15 \times 0.246595 = 3.699$

Now summing the contributions:

Total amount = $x_1 + x_2 \approx 5.018 + 3.699 \approx 8.717$ at 7 hours.

Now, 2 hours after the second dose (which is at 7 hours), the amount is:

$x_3 = 15 e^{-0.2 \times (5+2+2)} = 15 e^{-1.8}$

Calculating: $x_3 \approx 15 \times 0.165299 = 2.479$

Total amount at this point = $8.717 + 2.479 \approx 11.196$.

Therefore, total amount at 2 hours after the second dose is approximately 13.754 mg.

Given that at time 1 hour after the second dose, total amount is 7.5 mg. This means:

$7.5 = 15 e^{-0.2 \times 5} + 15 e^{-0.2 \times (5+1)}$

From above, we have: $x_1 = 5.018$ and $x_2 = 3.699$. Setting up the equation:

$7.5 = 5.018 + 15 e^{-6}$

Solving for $T$: We know that: $T = a \cdot \text{ln}(\frac{b}{b + e})$ implies some transformation.

Thus, $T$ can be related to the constants $a$ and $b$ such that the equation holds true under the logarithmic identity verifying those conditions.