Question

you are developing a model to predict the decay of a rare chemical substance. the amount of substance ( a(t) ), in milligrams (mg), remaining after ( t ) hours is modeled by the equation

( a(t)=a_{0} cdot e^{-lambda t} )

where ( a_{0} ) is the original amount of the substance.

a sample originally contains 250 mg of the substance. you need to determine how long it will take for only 80 mg to remain.

part a: write an equation to represent this situation and use logarithms to solve for ( t ). remember to show all of your work for solving the equation. round your answer to the nearest hundredth (3 points)

part b: explain why you need to use logarithms to solve this equation (1 point)

part c: will the sample ever get to exactly 0 mg? explain (2 points)

you may type your answers in the answer box. to type math, use ( sqrt{ } ). however, please remember that what you type must be mathematically correct, so i would recommend uploading a picture of your handwritten work using ( square ). if ( square ) does not seem to work for you, send pictures of your work to your teacher in a webmail. if you do not show work/explain your reasoning to justify your answers, you will not receive credit.

(6 points)

Explanation:

Part A

Step1: Substitute the given values into the formula

Given \(A_0 = 250\), \(A(t)=80\), the formula \(A(t)=A_0e^{-\lambda t}\) becomes \(80 = 250e^{-\lambda t}\). First, divide both sides by 250: \(\frac{80}{250}=e^{-\lambda t}\), so \(\frac{8}{25}=e^{-\lambda t}\).

Step2: Take the natural logarithm of both sides

Since \(y = e^x\) and \(x=\ln(y)\) are inverse - functions, taking the natural logarithm of both sides of \(\frac{8}{25}=e^{-\lambda t}\), we get \(\ln(\frac{8}{25})=-\lambda t\). Then \(t =-\frac{\ln(\frac{8}{25})}{\lambda}\). If we assume \(\lambda\) is known (in a typical radioactive - decay problem, if it was a half - life problem, we could find \(\lambda\) from \(A(t)=A_0(\frac{1}{2})^{\frac{t}{T}}\) and \(A(t) = A_0e^{-\lambda t}\), where \(T\) is the half - life and \(\lambda=\frac{\ln(2)}{T}\). But if we assume \(\lambda = 1\) (for the sake of solving the equation structure, since the problem only asks to set up and solve using logarithms in terms of the formula):
\(t=\ln(\frac{25}{8})\approx1.16\) (using a calculator, \(\ln(\frac{25}{8})=\ln(25)-\ln(8)\approx3.22 - 2.08 = 1.14\) (more accurately, \(\ln(25)=2\ln(5)\approx2\times1.61 = 3.22\), \(\ln(8) = 3\ln(2)\approx3\times0.69 = 2.07\), \(\ln(\frac{25}{8})\approx1.15\))

Part B

We use logarithms because the equation \(A = A_0e^{-\lambda t}\) is an exponential equation of the form \(y = a\cdot e^{bx}\). Logarithms are the inverse operation of exponentials. They allow us to isolate the variable \(t\) which is in the exponent. By taking the logarithm of both sides of the exponential equation, we can transform it into a linear - like equation (in terms of solving for \(t\)) using the property \(\ln(e^{x})=x\).

Part C

The sample will never reach exactly \(0\) mg. The function \(A(t)=A_0e^{-\lambda t}\) is an exponential decay function. As \(t
ightarrow\infty\), \(A(t)
ightarrow0\). Mathematically, \(e^{-\lambda t}=\frac{A(t)}{A_0}\). If \(A(t) = 0\), then \(e^{-\lambda t}=0\). But the exponential function \(y = e^{x}\) has a range of \((0,\infty)\) for all real values of \(x\). So, there is no real - value of \(t\) for which \(e^{-\lambda t}=0\).

Answer:

• Part A: The equation is \(80 = 250e^{-\lambda t}\), and \(t=\ln(\frac{25}{8})\approx1.15\) (assuming \(\lambda = 1\)).
• Part B: Logarithms are the inverse of exponentials, allowing us to solve for the variable in the exponent.
• Part C: No, because the exponential function \(y = e^{x}\) never equals \(0\) for real \(x\).