\(\text{www.the}\beta\text{etafunction.com}\)
Home
Login
Questions: Algebra BusinessCalculus
Please login to create an exam or a quiz.
The half life of a radioactive substance is 865 days. How log will it take until there is 49.5% of the substance remaining? Round your solution to the nearest tenth.
The decay constant is \(\displaystyle k = \frac{\ln 2}{865}\). This gives the equation \(\displaystyle 0.495 = e^{-\frac{\ln(2)}{865}t}\) Taking the natural logarithm of both sides gives \(\displaystyle \ln(0.495)= \frac{-t\ln(2)}{865}\). Solving for \(\displaystyle t\) gives \(\displaystyle t = -\frac{ 865\ln(0.495) }{ \ln(2) }\). It will take about about 877.5 days.
\begin{question}The half life of a radioactive substance is 865 days. How log will it take until there is 49.5\% of the substance remaining? Round your solution to the nearest tenth.
\soln{13.5cm}{The decay constant is $k = \frac{\ln 2}{865}$. This gives the equation $0.495 = e^{-\frac{\ln(2)}{865}t}$ Taking the natural logarithm of both sides gives $\ln(0.495)= \frac{-t\ln(2)}{865}$. Solving for $t$ gives $t = -\frac{ 865\ln(0.495) }{ \ln(2) }$. It will take about about 877.5 days. }
\end{question}
\documentclass{article}
\usepackage{tikz}
\usepackage{amsmath}
\usepackage[margin=2cm]{geometry}
\usepackage{tcolorbox}
\newcounter{ExamNumber}
\newcounter{questioncount}
\stepcounter{questioncount}
\newenvironment{question}{{\noindent\bfseries Question \arabic{questioncount}.}}{\stepcounter{questioncount}}
\renewcommand{\labelenumi}{{\bfseries (\alph{enumi})}}
\newif\ifShowSolution
\newcommand{\soln}[2]{%
\ifShowSolution%
\noindent\begin{tcolorbox}[colframe=blue,title=Solution]#2\end{tcolorbox}\else%
\vspace{#1}%
\fi%
}%
\newcommand{\hideifShowSolution}[1]{%
\ifShowSolution%
%
\else%
#1%
\fi%
}%
\everymath{\displaystyle}
\ShowSolutiontrue
\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>The half life of a radioactive substance is 865 days. How log will it take until there is 49.5% 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}{865} " src="/equation_images/%20%5Cdisplaystyle%20k%20%3D%20%5Cfrac%7B%5Cln%202%7D%7B865%7D%20" alt="LaTeX: \displaystyle k = \frac{\ln 2}{865} " data-equation-content=" \displaystyle k = \frac{\ln 2}{865} " /> . This gives the equation <img class="equation_image" title=" \displaystyle 0.495 = e^{-\frac{\ln(2)}{865}t} " src="/equation_images/%20%5Cdisplaystyle%200.495%20%3D%20e%5E%7B-%5Cfrac%7B%5Cln%282%29%7D%7B865%7Dt%7D%20" alt="LaTeX: \displaystyle 0.495 = e^{-\frac{\ln(2)}{865}t} " data-equation-content=" \displaystyle 0.495 = e^{-\frac{\ln(2)}{865}t} " /> Taking the natural logarithm of both sides gives <img class="equation_image" title=" \displaystyle \ln(0.495)= \frac{-t\ln(2)}{865} " src="/equation_images/%20%5Cdisplaystyle%20%5Cln%280.495%29%3D%20%5Cfrac%7B-t%5Cln%282%29%7D%7B865%7D%20" alt="LaTeX: \displaystyle \ln(0.495)= \frac{-t\ln(2)}{865} " data-equation-content=" \displaystyle \ln(0.495)= \frac{-t\ln(2)}{865} " /> . 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{ 865\ln(0.495) }{ \ln(2) } " src="/equation_images/%20%5Cdisplaystyle%20t%20%3D%20-%5Cfrac%7B%20865%5Cln%280.495%29%20%7D%7B%20%5Cln%282%29%20%7D%20" alt="LaTeX: \displaystyle t = -\frac{ 865\ln(0.495) }{ \ln(2) } " data-equation-content=" \displaystyle t = -\frac{ 865\ln(0.495) }{ \ln(2) } " /> . It will take about about 877.5 days. </p> </p>