At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg,...

Question:

At time t hours after taking the cough suppressant hydrocodone bitartrate, the amount, A, in mg, remaining in the body is given by {eq}\displaystyle A = 10(0.82)^t {/eq}.

(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 hours after the dose is administered?

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

Exponential Decay:

If a quantity decreases at a proportional rate to its present value, then we say that it undergoes exponential decay. The formula for the value of a quantity after {eq}t {/eq} years which undergoes exponential decay whose present value is {eq}P {/eq} with a rate {eq}r% {/eq} is:

$$A = P (1-r)^t $$

Answer and Explanation:

The given equation is:

$$A=10(0.82)^{t} = 10 (1-0.18)^t \,\,\, (\because 1-0.18 =0.82) $$

Comparing this with the equation of exponential decay:

$$A = P (1-r)^t $$

We get:

$$P= 10;\\ r=0.18 $$

(a) The initial amount taken is: {eq}P=\boxed{\mathbf{10}} {/eq}.


(b) The rate of decrease is:

$$r=0.18 $$

So the corresponding percentage is: {eq}0.18 \times 100 = \boxed{\mathbf{\text{18%}}} {/eq}.


(c) Substitute {eq}t=6 {/eq} in the given equation:

$$A = 10(0.82)^6 \approx \boxed{\mathbf{3.04 \text{ mg}}} $$


(d) It is given that {eq}A=1 {/eq}.

Substitute this in the given equation:

$$1=10(0.82)^t \\ 0.1 =0.82^t \\ \text{Apply natural logarithm on both sides}, \\ \ln 0.1 = \ln 0.82^t \\ \ln 0.1 = t \ln 0.82 \,\, (\because \ln a^m = m \ln a) \\ t = \dfrac{\ln 0.1}{\ln 0.82} \\ t \approx \boxed{\mathbf{11.6 \text{ hours}}} $$

Note: All answers are rounded to nearest tenth.


Learn more about this topic:

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Exponential Growth vs. Decay

from Math 101: College Algebra

Chapter 10 / Lesson 2
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