Question

The concentration of a drug in the body fluids is given by c=c0e⁽⁻⁰·⁸ᵗ⁾ , where c0 is the initial dosage and t is the time in hours elapsed after administering the dose. If 70mg of a drug is given, how much time elapses until 21mg of the drug remains?

Answer 1

The student's question involves calculating the time until the drug concentration decreases to a specific amount using an exponential decay formula. To find the time until only 21mg of a 70mg dose remains, logarithms are used to solve for variable t in the exponential equation.

Step-by-step explanation:

The student is asking for the time elapses until the drug concentration in the body fluids reduces from an initial dosage (c0) of 70mg to 21mg. Given the concentration formula c = c0e(-0.8t), we can use logarithms to solve for time (t).

Here's the calculation process:

1. First, substitute the known values into the formula: 21 = 70e(-0.8t).
2. Next, divide both sides by 70 to isolate the exponential expression: 21/70 = e(-0.8t).
3. Now, take the natural logarithm (ln) of both sides to solve for t: ln(21/70) = -0.8t.
4. Then, solve for t: t = ln(21/70) / -0.8.
5. After calculating, the time (t) when 21mg of the drug remains can be found.