Adrug 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(0.7)t and that in order for the drug to be effective, at least 4 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?

a - 4.512 hours

b - 1.94 hours

Step-by-step explanation:

Given,

a) A(t) = 10 (0.7)^t

To determine when 2mg is left in the body

We would have,

A(t) = 2, therefore

2 = 10(0.7)^t

0.7^t =2÷10

0.7^t = 0.2

Take the log of both sides,

Log (0.7)^t = log 0.2

t log 0.7 = log 0.2

t = log 0.2/ 0.7

t = 4.512 hours

Thus it will take 4.512 hours for 2mg to be left in the body.

b) Half life

Let A(t) = 1/2 A(0)

Thus,

1/2 A(0) = A(0)0.7^t

Divide both sides by A(0)

1/2 = 0.7^t

0.7^t = 0.5

Take log of both sides

Log 0.7^t = log 0.5

t log 0.7 = log 0.5

t = log 0.5/log 0.7

t = 1.94 hours

Therefore, the half life of the drug is 1.94 hours

It must have the same slope as the given line, so

m = -A/B = -1/-2 = 1/2

So far, we have

y = mx + b = x/2 + b

Use the given point to find b.

4 = -2/2 + b = -1 + b

b = 5

y = x/2 + 5

Multiply through by 2.

2y = x + 10

Standard form:

x - 2y = -10