# C3 2016 9 Edexcel

(a)

In this case, D = 15 so after 4 hours there are:

x = 15 e^{-0.2 \times 4} = 6.740 \text {mg (3dp)}

left after 4 hours.

(b) By this point, it has been 7 hours since the first dose was given so the amount remaining from the first dose is:

15e^{-0.2 \times 7} = 3.69895\ldots \text {mg}

The amount remaining of the second dose is:

15e^{-0.2 \times 2} = 10.0548\ldots \text {mg}

So the total amount of the antibiotic in the blood stream is 15e^{-0.2 \times 7} + 15e^{-0.2 \times 2} = 13.7538\ldots = 13.754 \text {mg (3dp)}

If it has been T hours since the second dose it has been T + 5 hours since the first dose.

So at T hours there is:

15e^{-0.2 (T + 5)} = 15e^{-0.2T - 1}

mg remaining of the first dose of antibiotic and:

15e^{-0.2 T} = 15e^{-0.2T}

So at time T we have:

15e^{-0.2 T} + 15e^{-0.2T - 1} = 7.5

So:

e^{-0.2 T} + e^{-0.2T - 1} = \dfrac 1 2

That is:

\displaystyle e^{-0.2T} + \frac 1 e \times e^{-0.2T} = \left(1 + \frac 1 e\right)e^{-0.2T} = \dfrac 1 2

So:

\displaystyle e^{-0.2T} = \dfrac 1 {2(1 + \frac 1 e)}

So:

\displaystyle -\frac 1 5 T = \ln \left(\frac 1 {2(1 + \frac 1 e)}\right)

Then:

\displaystyle T = 5 \ln \left(2 \left(1 + \frac 1 e\right)\right) = 5 \ln \left(2 + \frac 2 e\right)

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