A drug is eliminated from the body through urine. Suppose that for a dose of 10 milligrams, the amount A(t) remaining in the body t hours later is given by A(t)=10(0.8)^t and that in order for the drug to be effective, at least 2 milligrams must be in the body.
(a) Determine when 2 milligrams is left in the body.
(b) What is the half-life of the drug?
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To determine when 2 milligrams is left in the body, we need to find the time t when A(t) equals 2 milligrams.
(a) Setting A(t) equal to 2 milligrams and solving for t:
2 = 10(0.8)^t
To isolate the exponential term, divide both sides by 10:
0.2 = (0.8)^t
Now, take the logarithm (base 0.8) of both sides to solve for t:
log(base 0.8)(0.2) = t
Using the change of base formula, we can rewrite the equation:
log(0.2)/log(0.8) ≈ 6.9078/(-0.0969) ≈ -71.13
So, t ≈ -71.13 hours.
However, time cannot be negative in this context, so we can conclude that 2 milligrams is never left in the body.
(b) The half-life of a drug is the amount of time required for the drug to decrease to half of its original potency.
In this case, we need to find the time t when A(t) equals half the initial dose of 10 milligrams, which is 5 milligrams.
Setting A(t) equal to 5 milligrams and solving for t:
5 = 10(0.8)^t
Dividing both sides by 10:
0.5 = (0.8)^t
Taking the logarithm (base 0.8) of both sides:
log(base 0.8)(0.5) = t
Again, using the change of base formula:
log(0.5)/log(0.8) ≈ -0.3010/(-0.0969) ≈ 3.110
So, the half-life of the drug is approximately 3.110 hours.