Let C(t) be the concentration of a drug in the bloodstream. As the body eliminates the drug, C(t) decreases at a rate that is proportional to the amount of the drug that is present at the time. Thus C 0 (t) = −2kC(t), where 2k is a positive number. Here is it is half of the elimination constant of the drug. (a) If Co is the concentration at time t = 0, find the concentration at time . (b) If the body eliminates half the drug in 30 hours, how long does it take to eliminate 80% of the drug?

See explanation below

Step-by-step explanation:

In this case, let's answer this by parts:

a) Concentration at time t when Co  is concentration at t = 0:

In this case, we will use the initial expression but without the 2, so:

C(t) = -kC

In this case, we want to know the concentration at time t so:

C'(t) = -kC(t)    derivating:

dC/dt = -kC

dC/C = -kdt   From here, we can do integrals so:

lnC = -kt + C₁   (1)

Now, it's time to replace t = 0 and C = C₀:

lnC₀ = -k(0) + C₁

lnC₀ = C₁   (2)

Replacing (2) in (1) we have:

lnC = -kt + lnC₀

lnC - lnC₀ = -kt

ln(C/C₀) = -kt

C/C₀ = e^(-kt)

C(t) = C₀ e^(-kt)   (3)

This is the expression for C at given time t.

2. time to eliminate 80% of the drug:

With the first data, we need to calculate the value of k, which will be constant at any given time so:

C(t) = C₀ e^(-kt)

0.5C₀ = C₀ e^(-30k)

0.5 = e^(-30k)

ln(0.5) = -30k

k = 0.02310

Now with this value we can calculate the time to eliminate 80% of the drug or simply in other words, that we just have a remaining of 0.2C₀:

0.2C₀ = C₀ e^(-0.0231t)

ln(0.2) = -0.0231t

-ln(0.2) / 0.0231 = t

t = 69.67 h

This is the time to eliminate 80% of the drug

step-by-step explanation:

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b