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The half life of a radioactive substance is 1615 years. How log will it take until there is 34.6% of the substance remaining? Round your solution to the nearest tenth.
The decay constant is \(\displaystyle k = \frac{\ln 2}{1615}\). This gives the equation \(\displaystyle 0.346 = e^{-\frac{\ln(2)}{1615}t}\) Taking the natural logarithm of both sides gives \(\displaystyle \ln(0.346)= \frac{-t\ln(2)}{1615}\). Solving for \(\displaystyle t\) gives \(\displaystyle t = -\frac{ 1615\ln(0.346) }{ \ln(2) }\). It will take about about 2472.8 years.
\begin{question}The half life of a radioactive substance is 1615 years. How log will it take until there is 34.6\% of the substance remaining? Round your solution to the nearest tenth.
\soln{13.5cm}{The decay constant is $k = \frac{\ln 2}{1615}$. This gives the equation $0.346 = e^{-\frac{\ln(2)}{1615}t}$ Taking the natural logarithm of both sides gives $\ln(0.346)= \frac{-t\ln(2)}{1615}$. Solving for $t$ gives $t = -\frac{ 1615\ln(0.346) }{ \ln(2) }$. It will take about about 2472.8 years. }
\end{question}
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\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>The half life of a radioactive substance is 1615 years. How log will it take until there is 34.6% of the substance remaining? Round your solution to the nearest tenth.</p> </p>
<p> <p>The decay constant is <img class="equation_image" title=" \displaystyle k = \frac{\ln 2}{1615} " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B1615%7D%20" alt="LaTeX: \displaystyle k = \frac{\ln 2}{1615} " data-equation-content=" \displaystyle k = \frac{\ln 2}{1615} " /> . This gives the equation <img class="equation_image" title=" \displaystyle 0.346 = e^{-\frac{\ln(2)}{1615}t} " src="/equation_images/%20%5Cdisplaystyle%200.346%20%3D%20e%5E%7B-%5Cfrac%7B%5Cln%282%29%7D%7B1615%7Dt%7D%20" alt="LaTeX: \displaystyle 0.346 = e^{-\frac{\ln(2)}{1615}t} " data-equation-content=" \displaystyle 0.346 = e^{-\frac{\ln(2)}{1615}t} " /> Taking the natural logarithm of both sides gives <img class="equation_image" title=" \displaystyle \ln(0.346)= \frac{-t\ln(2)}{1615} " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%280.346%29%3D%20%5Cfrac%7B-t%5Cln%282%29%7D%7B1615%7D%20" alt="LaTeX: \displaystyle \ln(0.346)= \frac{-t\ln(2)}{1615} " data-equation-content=" \displaystyle \ln(0.346)= \frac{-t\ln(2)}{1615} " /> . 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{ 1615\ln(0.346) }{ \ln(2) } " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20-%5Cfrac%7B%201615%5Cln%280.346%29%20%7D%7B%20%5Cln%282%29%20%7D%20" alt="LaTeX: \displaystyle t = -\frac{ 1615\ln(0.346) }{ \ln(2) } " data-equation-content=" \displaystyle t = -\frac{ 1615\ln(0.346) }{ \ln(2) } " /> . It will take about about 2472.8 years. </p> </p>