At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount A, in mg, remaining in the body is given by A=13(0.82)^t.

a) What was the initial amount taken?
b) What percent of the drug leaves the body each hour?
c) How much of the drug is left in the body 6 hrs after the does is administered?
Round to two decimal places.
d) How long is it until only 1 mg of the drug remains in the body?
Round to two decimal places.

a) 13 mg

b) 18%
c) 13(.82)^6 = 3.95 g
d) 1 = 13(.82)^t so t=12.92 hrs

a) To find the initial amount taken, we can look at the equation A=13(0.82)^t. Since A represents the amount remaining in the body, we need to determine the initial amount before any time has passed.

At time t=0 hours, the equation becomes A=13(0.82)^0. Any number raised to the power of zero is equal to 1. Therefore, A=13(1), and it simplifies to A=13.

So, the initial amount taken is 13 mg.

b) The percent of the drug that leaves the body each hour can be determined by finding the change in amount over one hour. Since the equation already gives us the amount remaining, we can find the change by subtracting the amount at a given time from the amount at the previous time.

The amount remaining at time t+1 hour is given by A=13(0.82)^(t+1).

The amount remaining at time t hour is given by A=13(0.82)^t.

Change in amount = Initial amount - Remaining amount
Change in amount = 13(0.82)^t - 13(0.82)^(t+1)

To find the percentage, we need to divide the change in amount by the initial amount and multiply by 100.

Percent of drug leaving the body each hour = (Change in amount / Initial amount) x 100

Percent of drug leaving the body each hour = (13(0.82)^t - 13(0.82)^(t+1)) / 13 x 100

Simplifying further,

Percent of drug leaving the body each hour = (0.82^t - 0.82^(t+1)) x 100

Which gives us the final answer:

Percent of drug leaving the body each hour = 18%

c) To determine how much of the drug is left in the body 6 hours after the dose is administered, we substitute t=6 into the equation A=13(0.82)^t.

A = 13(0.82)^6

Using a calculator, we evaluate this:

A = 13(0.364)

A ≈ 4.73 mg

So, approximately 4.73 mg of the drug is left in the body 6 hours after the dose is administered.

d) To find out how long it takes until only 1 mg of the drug remains in the body, we can set the equation A=13(0.82)^t equal to 1 and solve for t.

13(0.82)^t = 1

Divide both sides of the equation by 13:

(0.82)^t = 1/13

Taking the logarithm of both sides with base 0.82:

t log(0.82) = log(1/13)

Divide both sides by log(0.82) to isolate t:

t = log(1/13) / log(0.82)

Using a calculator, we evaluate this:

t ≈ 8.8

So, it takes approximately 8.8 hours for only 1 mg of the drug to remain in the body.

To answer these questions, we will use the given equation A=13(0.82)^t, where A represents the amount of the drug remaining in the body at time t.

a) What was the initial amount taken?

The initial amount taken corresponds to the value of A when t = 0. Therefore, substitute t = 0 into the equation to find the initial amount taken:

A = 13(0.82)^0
A = 13(1)
A = 13 mg

So, the initial amount taken was 13 mg.

b) What percent of the drug leaves the body each hour?

To find the percent of the drug that leaves the body each hour, we need to determine how much of the drug remains after one hour. We can do this by calculating the difference between the initial amount taken and the amount remaining after one hour.

Let's calculate the amount remaining after one hour:

A(1) = 13(0.82)^1
A(1) = 13(0.82)
A(1) ≈ 10.66 mg

The difference between the initial amount and the amount remaining after one hour is:

13 - 10.66 = 2.34 mg

To find the percentage, divide this difference by the initial amount and multiply by 100:

(2.34/13) * 100 ≈ 18%

Therefore, approximately 18% of the drug leaves the body each hour.

c) How much of the drug is left in the body 6 hours after the dose is administered?

To find the amount remaining in the body 6 hours after the dose is administered, we substitute t = 6 into the equation and calculate:

A(6) = 13(0.82)^6
A(6) ≈ 5.16 mg

Therefore, approximately 5.16 mg of the drug is left in the body 6 hours after the dose is administered.

d) How long is it until only 1 mg of the drug remains in the body?

We need to find the value of t when A = 1. Substitute A = 1 into the equation and solve for t:

1 = 13(0.82)^t

Divide both sides by 13 to isolate the exponent:

(0.82)^t ≈ 1/13

Take the logarithm of both sides to solve for t:

log(0.82)^t ≈ log(1/13)

Using logarithm properties (log(b^x) = x*log(b)):

t * log(0.82) ≈ log(1/13)

Therefore, we can solve for t:

t ≈ log(1/13) / log(0.82)

Using a calculator, the approximate value of t is:

t ≈ 7.58 hours

Therefore, it takes approximately 7.58 hours until only 1 mg of the drug remains in the body.