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# Drug Stabiliser

It is not necessary to use algebra in these solutions but it often does help to consider how the numerical relationship may find a more general form of expression.

If 100mg of a drug is administered on Monday morning and it has a half life of 12 hours:

After 1 half life, the level of the drug would be $100 \times (\frac{1}{2})$ = 50mg

After 2 half lives, the level of the drug becomes $100 \times (\frac{1}{2}) \times (\frac{1}{2})$ = 25mg

After 3 half lives, the level of the drug becomes $100 \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2})$ = 12.5mg

After 4 half lives, the level of the drug is $100 \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2})$ = 6.25mg

This means that 4 half lives are needed before the 10mg level is passed. This is a period of 48 hours meaning Wednesday morning is the first morning that the level of drug drops below 10mg.

In a more general algebraic form, the level of drug falls according to the number of half lives, n. This value of n must give a level of drug such that:

$100 \times (\frac{1}{2})^{\text{n}}\leq 10$

$(\frac{1}{2})^{\text{n}}\leq 0.1$

$\text n\log_{10}(\frac{1}{2})\leq \log_{10} 0.1$

$\text n\geq \frac {-1}{\log_{10}(\frac{1}{2})}$

$\text n\geq 3.32...$

$\text {n}=4$(nearest whole number)

If another tablet is given on Wednesday morning, the residual level of 6.25mg is added to the 100mg consumed on the day to give 106.25mg.

The half life of the drug is 12 hours, thus there are $48\div{12}= 4$ half lives till Friday morning. This means that the level of drug in the body would be equal to:

$(106.25)\times(\frac{1}{2})^4 = {6.64}\text {mg}$ (3sf)

This calculation may also be approached stepwise:

106.25mg $\rightarrow$ 53.13mg $\rightarrow$ 26.56mg $\rightarrow$ 13.28mg $\rightarrow$ 6.64mg

It should be noted that each day comprises of 2 half lives and a dose of 100mg is given at the start of each morning. A stepwise approach yields:

100 $\times (\frac{1}{2})^2$ = 25 mg

(100 + 25) $\times(\frac{1}{2})^2$ = 31.25 mg

(100 + 31.25) $\times(\frac{1}{2})^2$ = 32.81 mg

(100 + 32.81) $\times(\frac{1}{2})^2$ = 33.20 mg

(100 + 33.20) $\times(\frac{1}{2})^2$ = 33.30 mg

(100 + 33.30) $\times(\frac{1}{2})^2$ = 33.33 mg

(100 + 33.33) $\times(\frac{1}{2})^2$ = 33.33 mg

...

The final minimum level through the iteration based method is 33.3 mg (3sf)

For an algebraic method, consider that the residual level at the start of the day would have to be once again attained after 2 half lives. Thus:

$(k + 100) \times (\frac{1}{2})^2) = k$

$\frac {(k+100)}{4} = k$

$k + 100 = 4k$

$3k = 100$

$k = \frac{100}{3}$

If a tablet is given every morning and every evening, there is only a period comprising 1 half life between doses. Therefore:

100$\times (\frac{1}{2})$ = 50 mg

(100 + 50) $\times(\frac{1}{2})$ = 75 mg

(100 + 75) $\times(\frac{1}{2})$ = 87.5 mg

(100 + 87.5) $\times(\frac{1}{2})$ = 93.75 mg

(100 + 93.75) $\times(\frac{1}{2})$ = 96.88 mg

(100 + 96.88) $\times(\frac{1}{2})$ = 98.44 mg

(100 + 98.44) $\times(\frac{1}{2})$ = 99.22 mg

...

This iterative method reveals an answer of 100 mg as the mnimum residual level.

An algebraic method considers that the minimum level is yielded again after the addition of a 100 mg dose after a period of 1 half life. So:

$\frac{(k + 100)}{2} = k$

$k + 100 = 2k$

$\therefore k = 100$

The half life, $\text t_\frac{1}{2}$= 6 days at the longest for an individual patient. If a dose of 20mg is given every day and a decay of $(\frac{1}{2})^\frac{1}{6}$is experience on a single day then the minimum level is given by a formula similar to those previously derived:

$(k + 20)\times(\frac{1}{2})^\frac{1}{6} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{6}$

$k(1- (\frac{1}{2})^\frac{-1}{6}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{6} -1}$

$k = 163.3$ mg (1dp)

Therefore the peak value is obtained by adding 20 mg as this is the dose given once this minimum level is obtained at the start of every day. This gives a value of 183.3 mg (1dp) as a peak level for an individual where it is the case that the drug has the longest possible half life.

If the calculation is repeated for t$_\frac{1}{2} = 4$ days, then the calculation above only requires a simple modification as a decay of $(\frac{1}{2})^{1}{4}$ is now experienced on a daily basis.

$(k + 20)\times(\frac{1}{2})^\frac{1}{4} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{4}$

$k(1- (\frac{1}{2})^\frac{-1}{4}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{4} -1}$

$k = 105.7$ mg (1dp) which gives a peak level of 125.7 mg (1dp).

An iterative method for the above problem may also be used. For example:

$20\times(\frac{1}{2})^\frac{1}{6} \rightarrow 17.82$

$(17.82 + 20)\times(\frac{1}{2})^\frac{1}{6}\rightarrow 33.69$

...

$(Ans + 20)\times(\frac{1}{2})^\frac{1}{6}$

Repeatedly computing this into a calculator will eventually yield the value of 163.3 mg as the minimum level. A smimlar approach may be utilised for the t$_\frac{1}{2} = 4$ days case .

This next part of the question is a little bit trickier, but consideration of relationships in a similar format to those already discussed provide a neat solution.

First consider the case where t$_\frac{1}{2} = 6$ days. We are looking for a situation where a dose given to a minimum level at the end of the week results in a peak value of 183.3 mg. If $\alpha$ = minimum residual drug level and $k$= weekly dose, then:

$\alpha + k = 183.3$

and

$(\alpha + k) (\frac{1}{2})^\frac{7}{6} = \alpha$

as the effective number of half lives during the week is $\frac{7}{6}$ and the residual level $\alpha$ must be obtained at the end of the week. Substitution of the the first form into the second gives the value of $\alpha$:

$183.3 \times (\frac{1}{2})^\frac{7}{6} = \alpha$

$\alpha = 81.6$ mg

$k = 183.3 - 81.6$

$k= 101.7$ mg (1dp)

for the case where t$_\frac{1}{2} = 4$, the calculation becomes:

$\alpha + k = 125.7$

$(\alpha + k) (\frac{1}{2})^\frac{7}{4} = \alpha$

$\alpha = 37.4$ mg

$k = 125.7 - 37.4$

$k = 88.3$ mg (1dp)

Any equivalent iterative methods are also valid, though an algebraic solution proves much easier to handle at this stage.

So in summary the lowest and highest long term levels of drug in the body are:

Daily

t$_\frac{1}{2} = 4$ days - high: 125.7 mg, low: 105.7 mg

t$_\frac{1}{2} = 6$ days - high: 183.3 mg, low: 163.3 mg

Weekly

t$_\frac{1}{2} = 4$ days - high: 125.7 mg, low: 37.4 mg

t$_\frac{1}{2} = 6$ days - high: 183.3 mg, low: 81.6 mg

The calculations additionally be attempted using t$_\frac{1}{2} = 5$ days which is an average of the two values given, though the insight given by using the upper and lower values of the half life is probably more insightful.

This raises several issues for a patient. A drug is a physiologically active substance and as such needs to be present in levels between doses that are sufficient to cause an effect. However drug levels in the body that are very high can cause unwanted side effects to become more pronounced. Add this to the fact that the half life of a drug in a patient can be highly variable and you can see that the situation is quite complex! Fluxoxetine is often marketed as Prozac and is an antidepressant approved for the treatment of major depression, obsessive compulsive disorder, bulimia and many other conditions. This example illustrates a case: drug level stability is an essential consideration as the underlying disorders the drug treats are highly serious and concern some kind of mental instability.

Thus if a patient has a weekly dose then if the half life of the drug in the body is 6 days then the peak and minimum levels calculated may be sensible depending on those required for the drug to be active. However, if the half life is 4 days then the long term minimal level may be too low! Consideration of daily dosing may mean that this option of drug delivery is not favourable for individuals where the half life of the drug is 6 days as persistently high levels of the drug may cause more manifest side effects to be expressed. This question is more open to intuitive interpretation of the data.

Drug companies would have to find a dosing level that would give a suitable peak drug level for the majority of their patients, taking into account the variability in half lives. As an interesting aside, consider that a drug company decides that dosing should occur every 3 days - will the patient remember when to take the drug for a period of a month? Companies often resort to producing packets of pills, a proportion of which don't contain the physiologically active compound to ensure the patient takes the pills regularly and obtains the correct dose.

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It is not necessary to use algebra in these solutions but it often does help to consider how the numerical relationship may find a more general form of expression.

If 100mg of a drug is administered on Monday morning and it has a half life of 12 hours:

After 1 half life, the level of the drug would be $100 \times (\frac{1}{2})$ = 50mg

After 2 half lives, the level of the drug becomes $100 \times (\frac{1}{2}) \times (\frac{1}{2})$ = 25mg

After 3 half lives, the level of the drug becomes $100 \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2})$ = 12.5mg

After 4 half lives, the level of the drug is $100 \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2}) \times (\frac{1}{2})$ = 6.25mg

This means that 4 half lives are needed before the 10mg level is passed. This is a period of 48 hours meaning Wednesday morning is the first morning that the level of drug drops below 10mg.

In a more general algebraic form, the level of drug falls according to the number of half lives, n. This value of n must give a level of drug such that:

$100 \times (\frac{1}{2})^{\text{n}}\leq 10$

$(\frac{1}{2})^{\text{n}}\leq 0.1$

$\text n\log_{10}(\frac{1}{2})\leq \log_{10} 0.1$

$\text n\geq \frac {-1}{\log_{10}(\frac{1}{2})}$

$\text n\geq 3.32...$

$\text {n}=4$(nearest whole number)

If another tablet is given on Wednesday morning, the residual level of 6.25mg is added to the 100mg consumed on the day to give 106.25mg.

The half life of the drug is 12 hours, thus there are $48\div{12}= 4$ half lives till Friday morning. This means that the level of drug in the body would be equal to:

$(106.25)\times(\frac{1}{2})^4 = {6.64}\text {mg}$ (3sf)

This calculation may also be approached stepwise:

106.25mg $\rightarrow$ 53.13mg $\rightarrow$ 26.56mg $\rightarrow$ 13.28mg $\rightarrow$ 6.64mg

It should be noted that each day comprises of 2 half lives and a dose of 100mg is given at the start of each morning. A stepwise approach yields:

100 $\times (\frac{1}{2})^2$ = 25 mg

(100 + 25) $\times(\frac{1}{2})^2$ = 31.25 mg

(100 + 31.25) $\times(\frac{1}{2})^2$ = 32.81 mg

(100 + 32.81) $\times(\frac{1}{2})^2$ = 33.20 mg

(100 + 33.20) $\times(\frac{1}{2})^2$ = 33.30 mg

(100 + 33.30) $\times(\frac{1}{2})^2$ = 33.33 mg

(100 + 33.33) $\times(\frac{1}{2})^2$ = 33.33 mg

...

The final minimum level through the iteration based method is 33.3 mg (3sf)

For an algebraic method, consider that the residual level at the start of the day would have to be once again attained after 2 half lives. Thus:

$(k + 100) \times (\frac{1}{2})^2) = k$

$\frac {(k+100)}{4} = k$

$k + 100 = 4k$

$3k = 100$

$k = \frac{100}{3}$

If a tablet is given every morning and every evening, there is only a period comprising 1 half life between doses. Therefore:

100$\times (\frac{1}{2})$ = 50 mg

(100 + 50) $\times(\frac{1}{2})$ = 75 mg

(100 + 75) $\times(\frac{1}{2})$ = 87.5 mg

(100 + 87.5) $\times(\frac{1}{2})$ = 93.75 mg

(100 + 93.75) $\times(\frac{1}{2})$ = 96.88 mg

(100 + 96.88) $\times(\frac{1}{2})$ = 98.44 mg

(100 + 98.44) $\times(\frac{1}{2})$ = 99.22 mg

...

This iterative method reveals an answer of 100 mg as the mnimum residual level.

An algebraic method considers that the minimum level is yielded again after the addition of a 100 mg dose after a period of 1 half life. So:

$\frac{(k + 100)}{2} = k$

$k + 100 = 2k$

$\therefore k = 100$

The half life, $\text t_\frac{1}{2}$= 6 days at the longest for an individual patient. If a dose of 20mg is given every day and a decay of $(\frac{1}{2})^\frac{1}{6}$is experience on a single day then the minimum level is given by a formula similar to those previously derived:

$(k + 20)\times(\frac{1}{2})^\frac{1}{6} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{6}$

$k(1- (\frac{1}{2})^\frac{-1}{6}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{6} -1}$

$k = 163.3$ mg (1dp)

Therefore the peak value is obtained by adding 20 mg as this is the dose given once this minimum level is obtained at the start of every day. This gives a value of 183.3 mg (1dp) as a peak level for an individual where it is the case that the drug has the longest possible half life.

If the calculation is repeated for t$_\frac{1}{2} = 4$ days, then the calculation above only requires a simple modification as a decay of $(\frac{1}{2})^{1}{4}$ is now experienced on a daily basis.

$(k + 20)\times(\frac{1}{2})^\frac{1}{4} = k$

$k + 20 = k (\frac{1}{2})^\frac{-1}{4}$

$k(1- (\frac{1}{2})^\frac{-1}{4}) = -20$

$k = \frac{20}{(\frac{1}{2})^\frac{-1}{4} -1}$

$k = 105.7$ mg (1dp) which gives a peak level of 125.7 mg (1dp).

An iterative method for the above problem may also be used. For example:

$20\times(\frac{1}{2})^\frac{1}{6} \rightarrow 17.82$

$(17.82 + 20)\times(\frac{1}{2})^\frac{1}{6}\rightarrow 33.69$

...

$(Ans + 20)\times(\frac{1}{2})^\frac{1}{6}$

Repeatedly computing this into a calculator will eventually yield the value of 163.3 mg as the minimum level. A smimlar approach may be utilised for the t$_\frac{1}{2} = 4$ days case .

This next part of the question is a little bit trickier, but consideration of relationships in a similar format to those already discussed provide a neat solution.

First consider the case where t$_\frac{1}{2} = 6$ days. We are looking for a situation where a dose given to a minimum level at the end of the week results in a peak value of 183.3 mg. If $\alpha$ = minimum residual drug level and $k$= weekly dose, then:

$\alpha + k = 183.3$

and

$(\alpha + k) (\frac{1}{2})^\frac{7}{6} = \alpha$

as the effective number of half lives during the week is $\frac{7}{6}$ and the residual level $\alpha$ must be obtained at the end of the week. Substitution of the the first form into the second gives the value of $\alpha$:

$183.3 \times (\frac{1}{2})^\frac{7}{6} = \alpha$

$\alpha = 81.6$ mg

$k = 183.3 - 81.6$

$k= 101.7$ mg (1dp)

for the case where t$_\frac{1}{2} = 4$, the calculation becomes:

$\alpha + k = 125.7$

$(\alpha + k) (\frac{1}{2})^\frac{7}{4} = \alpha$

$\alpha = 37.4$ mg

$k = 125.7 - 37.4$

$k = 88.3$ mg (1dp)

Any equivalent iterative methods are also valid, though an algebraic solution proves much easier to handle at this stage.

So in summary the lowest and highest long term levels of drug in the body are:

Daily

t$_\frac{1}{2} = 4$ days - high: 125.7 mg, low: 105.7 mg

t$_\frac{1}{2} = 6$ days - high: 183.3 mg, low: 163.3 mg

Weekly

t$_\frac{1}{2} = 4$ days - high: 125.7 mg, low: 37.4 mg

t$_\frac{1}{2} = 6$ days - high: 183.3 mg, low: 81.6 mg

The calculations additionally be attempted using t$_\frac{1}{2} = 5$ days which is an average of the two values given, though the insight given by using the upper and lower values of the half life is probably more insightful.

This raises several issues for a patient. A drug is a physiologically active substance and as such needs to be present in levels between doses that are sufficient to cause an effect. However drug levels in the body that are very high can cause unwanted side effects to become more pronounced. Add this to the fact that the half life of a drug in a patient can be highly variable and you can see that the situation is quite complex! Fluxoxetine is often marketed as Prozac and is an antidepressant approved for the treatment of major depression, obsessive compulsive disorder, bulimia and many other conditions. This example illustrates a case: drug level stability is an essential consideration as the underlying disorders the drug treats are highly serious and concern some kind of mental instability.

Thus if a patient has a weekly dose then if the half life of the drug in the body is 6 days then the peak and minimum levels calculated may be sensible depending on those required for the drug to be active. However, if the half life is 4 days then the long term minimal level may be too low! Consideration of daily dosing may mean that this option of drug delivery is not favourable for individuals where the half life of the drug is 6 days as persistently high levels of the drug may cause more manifest side effects to be expressed. This question is more open to intuitive interpretation of the data.

Drug companies would have to find a dosing level that would give a suitable peak drug level for the majority of their patients, taking into account the variability in half lives. As an interesting aside, consider that a drug company decides that dosing should occur every 3 days - will the patient remember when to take the drug for a period of a month? Companies often resort to producing packets of pills, a proportion of which don't contain the physiologically active compound to ensure the patient takes the pills regularly and obtains the correct dose.