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 Homework Statement:

A single dose of 10 mg of the drug with 200 ml water immediately after a low fat meal consisting of fruit juice, skim milk, cereal and toast with jelly. The concentration of the drug (in units of ##\mu##g ##l^{1}##) in the blood following treatment A is given by the following equation:
[tex] C_p (t) = 91.417 \left( e^{0.4t}  e^{3.455t} \right) [/tex]
Knowing that the minimal effective blood concentration (MEC) of the drug is 50 ##\mu##g per litre blood, calculate the duration of action for treatment A assuming that the time it takes to initially get above the MEC is 20 minutes.
 Relevant Equations:
 Concentration
Hi,
I was attempting the following question, but didn't understand how to make further progress with it.
Question:
A single dose of 10 mg of the drug with 200 ml water immediately after a low fat meal consisting of fruit juice, skim milk, cereal and toast with jelly. The concentration of the drug (in units of ##\mu##g ##l^{1}##) in the blood following treatment A is given by the following equation:
[tex] C_p (t) = 91.417 \left( e^{0.4t}  e^{3.455t} \right) [/tex]
Knowing that the minimal effective blood concentration (MEC) of the drug is 50 ##\mu##g per litre blood, calculate the duration of action for treatment A assuming that the time it takes to initially get above the MEC is 20 minutes.
Attempt:
1) What is the duration of action?
Is it the time which the drug is effective? In this case, that would be when the concentration is above 50 ##\mu##g per liter of blood.
2) How can I solve this question by hand?
This problem was set to be done without the use of a computer or advanced calculator techniques...
If I let ## C_p (t) = 50 ##, we get:
[tex] 50 = 91.417 \left( e^{0.4t}  e^{3.455t} \right) [/tex]
which is nonlinear and I don't understand how to proceed. Perhaps we can make an assumption that the second term is significantly smaller than the first by now  thus we can ignore it to yield:
[tex] 50 \approx 91.417 \left( e^{0.4t} \right) \rightarrow T = ln(\frac{50}{91.417})/0.4 = 1.508... \text{hrs} [/tex]
Then we can subtract the time taken for the drug to first become effective: ## \text{Duration} = 1.508  1/3 = 1.17518... \text{hours} ##
Does this approach seem correct?
I was attempting the following question, but didn't understand how to make further progress with it.
Question:
A single dose of 10 mg of the drug with 200 ml water immediately after a low fat meal consisting of fruit juice, skim milk, cereal and toast with jelly. The concentration of the drug (in units of ##\mu##g ##l^{1}##) in the blood following treatment A is given by the following equation:
[tex] C_p (t) = 91.417 \left( e^{0.4t}  e^{3.455t} \right) [/tex]
Knowing that the minimal effective blood concentration (MEC) of the drug is 50 ##\mu##g per litre blood, calculate the duration of action for treatment A assuming that the time it takes to initially get above the MEC is 20 minutes.
Attempt:
1) What is the duration of action?
Is it the time which the drug is effective? In this case, that would be when the concentration is above 50 ##\mu##g per liter of blood.
2) How can I solve this question by hand?
This problem was set to be done without the use of a computer or advanced calculator techniques...
If I let ## C_p (t) = 50 ##, we get:
[tex] 50 = 91.417 \left( e^{0.4t}  e^{3.455t} \right) [/tex]
which is nonlinear and I don't understand how to proceed. Perhaps we can make an assumption that the second term is significantly smaller than the first by now  thus we can ignore it to yield:
[tex] 50 \approx 91.417 \left( e^{0.4t} \right) \rightarrow T = ln(\frac{50}{91.417})/0.4 = 1.508... \text{hrs} [/tex]
Then we can subtract the time taken for the drug to first become effective: ## \text{Duration} = 1.508  1/3 = 1.17518... \text{hours} ##
Does this approach seem correct?