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Questions: Algebra BusinessCalculus
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After the release of radioactive material into the atmosphere from a nuclear power plant the hay in that country was contaminated by a radioactive isotope (half-life 54 weeks). If it is safe to feed the hay to cows when 25% of the radioactive isotope remains, how long did the farmers need to wait to use this hay? Round to the nearest tenth.
The decay constant is \(\displaystyle k = \frac{\ln 2}{54}\). This gives the equation \(\displaystyle 0.25 = e^{-\frac{\ln(2)}{54}t}\) Taking the natural logarithm of both sides gives \(\displaystyle \ln(0.25)= \frac{-t\ln(2)}{54}\). Solving for \(\displaystyle t\) gives \(\displaystyle t = -\frac{ 54\ln(0.25) }{ \ln(2) }\). The farmers had to wait about 108.0000000 weeks.
\begin{question}After the release of radioactive material into the atmosphere from a nuclear power plant the hay in that country was contaminated by a radioactive isotope (half-life 54 weeks). If it is safe to feed the hay to cows when 25% of the radioactive isotope remains, how long did the farmers need to wait to use this hay? Round to the nearest tenth.
\soln{4.5cm}{The decay constant is $k = \frac{\ln 2}{54}$. This gives the equation $0.25 = e^{-\frac{\ln(2)}{54}t}$ Taking the natural logarithm of both sides gives $\ln(0.25)= \frac{-t\ln(2)}{54}$. Solving for $t$ gives $t = -\frac{ 54\ln(0.25) }{ \ln(2) }$. The farmers had to wait about 108.0000000 weeks. }
\end{question}
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\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>After the release of radioactive material into the atmosphere from a nuclear power plant the hay in that country was contaminated by a radioactive isotope (half-life 54 weeks). If it is safe to feed the hay to cows when 25% of the radioactive isotope remains, how long did the farmers need to wait to use this hay? Round to the nearest tenth. </p> </p>
<p> <p>The decay constant is <img class="equation_image" title=" \displaystyle k = \frac{\ln 2}{54} " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B54%7D%20" alt="LaTeX: \displaystyle k = \frac{\ln 2}{54} " data-equation-content=" \displaystyle k = \frac{\ln 2}{54} " /> . This gives the equation <img class="equation_image" title=" \displaystyle 0.25 = e^{-\frac{\ln(2)}{54}t} " src="/equation_images/%20%5Cdisplaystyle%200.25%20%3D%20e%5E%7B-%5Cfrac%7B%5Cln%282%29%7D%7B54%7Dt%7D%20" alt="LaTeX: \displaystyle 0.25 = e^{-\frac{\ln(2)}{54}t} " data-equation-content=" \displaystyle 0.25 = e^{-\frac{\ln(2)}{54}t} " /> Taking the natural logarithm of both sides gives <img class="equation_image" title=" \displaystyle \ln(0.25)= \frac{-t\ln(2)}{54} " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%280.25%29%3D%20%5Cfrac%7B-t%5Cln%282%29%7D%7B54%7D%20" alt="LaTeX: \displaystyle \ln(0.25)= \frac{-t\ln(2)}{54} " data-equation-content=" \displaystyle \ln(0.25)= \frac{-t\ln(2)}{54} " /> . Solving for <img class="equation_image" title=" \displaystyle t " src="/equation_images/%20%5Cdisplaystyle%20t%20" alt="LaTeX: \displaystyle t " data-equation-content=" \displaystyle t " /> gives <img class="equation_image" title=" \displaystyle t = -\frac{ 54\ln(0.25) }{ \ln(2) } " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20-%5Cfrac%7B%2054%5Cln%280.25%29%20%7D%7B%20%5Cln%282%29%20%7D%20" alt="LaTeX: \displaystyle t = -\frac{ 54\ln(0.25) }{ \ln(2) } " data-equation-content=" \displaystyle t = -\frac{ 54\ln(0.25) }{ \ln(2) } " /> . The farmers had to wait about 108.0000000 weeks. </p> </p>