Question

score on last try: 13.33 of 20 pts. see details for more.
at least one scored part is incorrect. jump to first changable incorrect part.
get a similar question you can retry this question below
caffeine has a half - life of 3.9 hours. an average 8 ounce cup of tea contains 27 milligrams (mg) of caffeine.
suppose you consume an average 8 ounce cup of tea.
use the exponential decay model
$a = a_0e^{kt}$
where t is measured in hours, to write an exponential decay function for amount (milligrams) of caffeine in your body, t hours after consuming the cup of tea. round the decay constant, k to four significant figures.
$a(t)=27e^{- 0.1777t}$
after 20 hours, how many milligrams of caffeine will be left in your body? round the solution to one decimal place, if necessary.
0.8 milligrams
assuming no additional caffeine is consumed, how many hours will it take for the amount of caffeine in your body to reduce to 0.6 mg? round the solution to the nearest whole number.
21 hours

Explanation:

Step1: Recall half - life formula for exponential decay

For exponential decay, the half - life formula is related to the decay constant \(k\) by \(A = A_0e^{kt}\), and when \(t = T_{1/2}\) (half - life), \(A=\frac{A_0}{2}\). So we have \(\frac{A_0}{2}=A_0e^{kT_{1/2}}\).
We can cancel out \(A_0\) (since \(A_0 eq0\)) and get \(\frac{1}{2}=e^{kT_{1/2}}\).
Given \(T_{1/2} = 3.9\) hours, take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{k\times3.9})\).
Using the property \(\ln(e^x)=x\), we have \(\ln(\frac{1}{2}) = 3.9k\).

Step2: Solve for \(k\)

We know that \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.6931\).
So \(k=\frac{\ln(\frac{1}{2})}{3.9}=\frac{- 0.6931}{3.9}\approx - 0.1777\)? Wait, no, wait. Wait, the formula for exponential decay with half - life can also be written as \(A = A_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), and if we want to write it in the form \(A = A_0e^{kt}\), we can convert \((\frac{1}{2})^{\frac{t}{T_{1/2}}}\) to exponential form.
\((\frac{1}{2})^{\frac{t}{T_{1/2}}}=e^{\ln((\frac{1}{2})^{\frac{t}{T_{1/2}}})}=e^{\frac{t}{T_{1/2}}\ln(\frac{1}{2})}=e^{t\times\frac{\ln(\frac{1}{2})}{T_{1/2}}}\)
So \(k = \frac{\ln(\frac{1}{2})}{T_{1/2}}\). Let's calculate \(k\) correctly.
\(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693147\)
\(k=\frac{- 0.693147}{3.9}\approx - 0.17773\)
Wait, but the initial function was written as \(A(t)=27e^{- 0.1777t}\), but let's check again. Wait, maybe there was a miscalculation. Wait, let's recalculate \(k\):
\(k=\frac{\ln(1/2)}{3.9}=\frac{- \ln(2)}{3.9}\approx\frac{- 0.693147}{3.9}\approx - 0.1777\) (to four significant figures). Wait, but maybe the error is in the first part. Wait, the problem says "exponential decay model \(A = A_0e^{kt}\)". Let's check the first part again.
Wait, \(A_0 = 27\) (initial amount of caffeine). The half - life is \(t = 3.9\) hours, so when \(t = 3.9\), \(A=\frac{27}{2}\).
So \(\frac{27}{2}=27e^{k\times3.9}\)
Divide both sides by 27: \(\frac{1}{2}=e^{3.9k}\)
Take natural log: \(\ln(1/2)=3.9k\)
\(k=\frac{\ln(1/2)}{3.9}=\frac{- \ln(2)}{3.9}\approx\frac{- 0.6931}{3.9}\approx - 0.1777\). Wait, but maybe the user made a mistake in the first part? Wait, no, the first part's answer was marked wrong. Wait, no, wait, maybe the formula is \(A(t)=27e^{- 0.1777t}\)? Wait, let's check the calculation of \(k\) again.
\(\ln(2)\approx0.693147\), so \(\frac{\ln(2)}{3.9}\approx\frac{0.693147}{3.9}\approx0.1777\), but since it's decay, \(k\) should be negative. So \(k =-\frac{\ln(2)}{3.9}\approx - 0.1777\). Wait, but maybe the system expects a different approach? Wait, no, let's check the first part again. Wait, the user's first answer was \(A(t)=27e^{- 0.1777t}\), but maybe it's a formatting error? Wait, no, maybe the correct \(k\) is calculated as follows:
Wait, let's do the calculation more accurately. \(\ln(2)=0.69314718056\)
\(k =-\frac{0.69314718056}{3.9}\approx - 0.17773\). So \(k\approx - 0.1777\) (four significant figures). Wait, but maybe the problem is that the user wrote \(A(t)=27e^{- 0.1777t}\) but the system marked it wrong? Wait, no, maybe I made a mistake. Wait, let's check the half - life formula again. The general formula for exponential decay is \(A(t)=A_0e^{-kt}\), where \(k>0\). And the relationship between \(k\) and half - life \(T\) is \(k=\frac{\ln(2)}{T}\). Wait, yes! I had a sign error earlier. Because if \(A(t)=A_0e^{-kt}\), then when \(t = T\), \(A(T)=\frac{A_0}{2}\), so \(\frac{A_0}{2}=A_0e^{-kT}\), then \(\frac{1}{2}=e^{-kT}\), take natural log: \(\ln(\frac{1}{2})=-kT\), so \(k = \frac{\ln(2)}{T}\). Oh! That's the mistake. I had the sign wrong. So \(k=\frac{\ln(2)}{3.9}\approx\frac{0.6931}{3.9}\approx0.1777\), but since the formula is \(A(t)=A_0e^{-kt}\), then \(k\) is positive, and the exponent is \(-kt\). So \(A(t)=27e^{- 0.1777t}\) is correct? Wait, no, let's re - derive:
Start with \(A(t)=A_0e^{kt}\), for decay, \(k<0\). Let \(k=-r\), \(r > 0\). Then \(A(t)=A_0e^{-rt}\). When \(t = T_{1/2}\), \(A=\frac{A_0}{2}\), so \(\frac{A_0}{2}=A_0e^{-rT_{1/2}}\), so \(\frac{1}{2}=e^{-rT_{1/2}}\), \(\ln(\frac{1}{2})=-rT_{1/2}\), \(r=\frac{\ln(2)}{T_{1/2}}\), so \(k=-r =-\frac{\ln(2)}{T_{1/2}}\). So \(k =-\frac{\ln(2)}{3.9}\approx - 0.1777\). So the function \(A(t)=27e^{- 0.1777t}\) is correct. Wait, but the system marked it wrong. Maybe it's a formatting issue, like the exponent should be written as \(A(t)=27e^{(-0.1777t)}\) or something. But let's proceed to the first part correction.

Wait, the first part is to write the exponential decay function. \(A_0 = 27\), \(k=-\frac{\ln(2)}{3.9}\approx - 0.1777\), so \(A(t)=27e^{- 0.1777t}\) is correct. Maybe the system had a display error. Now, for the first part, the correct function is \(A(t)=27e^{- 0.1777t}\) (wait, but let's check the calculation of \(k\) again with more precision. \(\ln(2)=0.69314718056\), \(0.69314718056\div3.9 = 0.17773\), so \(k=- 0.17773\), so \(k\approx - 0.1777\) (four significant figures). So the function is correct.

Now, for the second part, after 20 hours, \(t = 20\), \(A(20)=27e^{- 0.1777\times20}\). Let's calculate the exponent: \(-0.1777\times20=-3.554\). Then \(e^{-3.554}\approx0.0284\). Then \(A(20)=27\times0.0284\approx0.7668\approx0.8\) (rounded to one decimal place), which matches the user's answer.

For the third part, we want to find \(t\) when \(A(t) = 0.6\). So \(0.6=27e^{- 0.1777t}\). Divide both sides by 27: \(\frac{0.6}{27}=e^{- 0.1777t}\), \(\frac{1}{45}=e^{- 0.1777t}\) (simplify \(\frac{0.6}{27}=\frac{6}{270}=\frac{1}{45}\approx0.0222\)). Take natural log: \(\ln(\frac{1}{45})=- 0.1777t\). \(\ln(\frac{1}{45})=-\ln(45)\approx - 3.8067\). So \(t=\frac{\ln(45)}{0.1777}\approx\frac{3.8067}{0.1777}\approx21.4\approx21\) (rounded to the nearest whole number), which matches the user's answer.

Wait, but the first part was marked wrong. Let's check the calculation of \(k\) again. Wait, maybe the formula is \(A(t)=27(\frac{1}{2})^{\frac{t}{3.9}}\), and if we convert it to \(A(t)=27e^{kt}\), we can do it as follows: \((\frac{1}{2})^{\frac{t}{3.9}}=e^{\ln((\frac{1}{2})^{\frac{t}{3.9}})}=e^{\frac{t}{3.9}\ln(\frac{1}{2})}=e^{t\times\frac{\ln(\frac{1}{2})}{3.9}}=e^{t\times(-\frac{\ln(2)}{3.9})}\), so \(k =-\frac{\ln(2)}{3.9}\approx - 0.1777\), so the function \(A(t)=27e^{- 0.1777t}\) is correct. Maybe the system had a typo or a formatting issue.

Answer:

The correct exponential decay function is \(A(t) = 27e^{-0.1777t}\) (the previous marking might be incorrect due to formatting or system error), after 20 hours, the amount of caffeine left is \(0.8\) milligrams, and the time to reduce to \(0.6\) mg is \(21\) hours.