To determine how long the farmers need to wait before feeding the hay to the cows safely, we can use the concept of radioactive decay. The radioactive decay formula is:
[tex]\[ N(t) = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T}} \][/tex]
Where:
- [tex]\( N(t) \)[/tex] is the final amount of radioactive material.
- [tex]\( N_0 \)[/tex] is the initial amount of radioactive material.
- [tex]\( t \)[/tex] is the time that has passed.
- [tex]\( T \)[/tex] is the half-life of the isotope.
Given the information:
- The initial amount ([tex]\( N_0 \)[/tex]) of the radioactive material is 100%, so we can set [tex]\( N_0 = 1 \)[/tex].
- The final amount ([tex]\( N(t) \)[/tex]) of the radioactive material is 9%, so we set [tex]\( N(t) = 0.09 \)[/tex].
- The half-life ([tex]\( T \)[/tex]) of the radioactive isotope is 30 years.
We need to find the time [tex]\( t \)[/tex] it takes for the amount of radioactive material to decay to 9% of its original amount.
The formula can be rearranged to solve for [tex]\( t \)[/tex]:
[tex]\[ \frac{N(t)}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T}} \][/tex]
Take the natural logarithm (ln) on both sides of the equation:
[tex]\[ \ln\left(\frac{N(t)}{N_0}\right) = \ln\left(\left(\frac{1}{2}\right)^{\frac{t}{T}}\right) \][/tex]
Using the properties of logarithms, we can simplify the right-hand side:
[tex]\[ \ln\left(\frac{N(t)}{N_0}\right) = \frac{t}{T} \ln\left(\frac{1}{2}\right) \][/tex]
Now isolate [tex]\( t \)[/tex]:
[tex]\[ t = \frac{\ln\left(\frac{N(t)}{N_0}\right)}{\ln\left(\frac{1}{2}\right)} \cdot T \][/tex]
Substitute the given values [tex]\( N(t) = 0.09 \)[/tex], [tex]\( N_0 = 1 \)[/tex], and [tex]\( T = 30 \)[/tex] years:
[tex]\[ t = \frac{\ln(0.09)}{\ln(0.5)} \cdot 30 \][/tex]
This calculation gives:
[tex]\[ t \approx 104.2 \][/tex]
Thus, the farmers needed to wait approximately 104.2 days for it to be safe to feed the hay to the cows. Rounded to one decimal place, the answer is:
The farmers needed to wait approximately 104.2 days for it to be safe to feed the hay to the cows.
