- Set the equation \(\frac{3}{5} N_0 = N_0 e^{-\frac{t \ln 2}{4.8}}\) using the given initial conditions.
- Take the natural logarithm of both sides to isolate the term with \(t\): \(\ln(\frac{3}{5}) = -\frac{t \ln 2}{4.8}\).
- Rearrange to solve for \(t\), resulting in \(t = 4.8 \frac{\ln(\frac{5}{3})}{\ln 2}\) or \(t = 4.8 \frac{\ln(\frac{3}{5})}{\ln(\frac{1}{2})}\).
- Based on the given options, the final answer is \(t = 4.8 \left( \frac{\ln \frac{3}{5}}{\ln \frac{1}{2}} \right)\).
Explanation
- Understanding the Problem and Data Analysis
We are given the half-life formula \(N(t) = N_0 e^{-\frac{t \ln 2}{h}}\), where \(N_0\) is the initial amount of a substance, \(h\) is its half-life, and \(N(t)\) is the amount remaining at time \(t\). We are given that the half-life of caffeine is \(h = 4.8\) hours, and the initial amount of caffeine in a cup of coffee is \(N_0 = 18\) ounces. We want to find the time \(t\) at which only \(60\%\) of the initial amount of caffeine remains, which means \(N(t) = 0.60 \times N_0\).
- Setting up the Equation
Substituting the given values into the formula:
\[0.60 N_0 = N_0 e^{-\frac{t \ln 2}{4.8}}\]
Dividing both sides by the initial amount \(N_0\) (which is non-zero):
\[0.60 = e^{-\frac{t \ln 2}{4.8}}\]
We know that \(0.60 = \frac{60}{100} = \frac{3}{5}\), so the equation is:
\[\frac{3}{5} = e^{-\frac{t \ln 2}{4.8}}\]
- Solving for Time t
To solve for \(t\), we take the natural logarithm (\(\ln\)) of both sides:
\[\ln\left(\frac{3}{5}\right) = \ln\left(e^{-\frac{t \ln 2}{4.8}}\right)\]
Using the property \(\ln(e^x) = x\), we get:
\[\ln\left(\frac{3}{5}\right) = -\frac{t \ln 2}{4.8}\]
Multiplying both sides by \(-4.8\) and dividing by \(\ln 2\):
\[t = -4.8 \frac{\ln\left(\frac{3}{5}\right)}{\ln 2}\]
Using the logarithm property \(a \ln(b) = \ln(b^a)\), we can write the negative sign as an exponent:
\[t = 4.8 \frac{-\ln\left(\frac{3}{5}\right)}{\ln 2} = 4.8 \frac{\ln\left(\left(\frac{3}{5}\right)^{-1}\right)}{\ln 2} = 4.8 \frac{\ln\left(\frac{5}{3}\right)}{\ln 2}\]
Alternatively, we can use the property \(\ln(1/2) = \ln(2^{-1}) = -\ln 2\), meaning \(\frac{1}{-\ln 2} = \frac{1}{\ln(1/2)}\). Therefore:
\[t = 4.8 \frac{\ln\left(\frac{3}{5}\right)}{\ln\left(\frac{1}{2}\right)}\]
Examples
Understanding half-life is crucial in various fields. For example, in medicine, pharmacists use half-life to determine how frequently a patient should take a specific medication to maintain an effective, safe level in their bloodstream. In environmental science, researchers use the half-life of radioactive pollutants to predict how long it will take for a contaminated area to become safe for human habitation. In our case, knowing the half-life of caffeine helps predict how long it will take for the effects of a coffee to wear off.