Question

a certain isotope decays so that the amount a remaining after t years is given by:
$a = a_0cdot e^{-0.03t}$ where $a_0$ is the original amount of the isotope.
to the nearest year, the half - life of the isotope (the amount of time it takes to decay to half the original amount) is
years.

Explanation:

Step1: Set up the equation

Since the amount \(A\) is half of the original amount \(A_0\), we set \(A=\frac{A_0}{2}\). Substituting into the formula \(A = A_0\cdot e^{- 0.03t}\), we get \(\frac{A_0}{2}=A_0\cdot e^{-0.03t}\). Divide both sides by \(A_0\) (since \(A_0
eq0\)), the equation becomes \(\frac{1}{2}=e^{-0.03t}\).

Step2: Take the natural logarithm of both sides

Using the property \(\ln(a^b)=b\ln(a)\) and \(\ln(e) = 1\), for the equation \(\frac{1}{2}=e^{-0.03t}\), we take the natural logarithm of both sides: \(\ln(\frac{1}{2})=\ln(e^{-0.03t})\). So, \(\ln(\frac{1}{2})=- 0.03t\).

Step3: Solve for \(t\)

We know that \(\ln(\frac{1}{2})=-\ln(2)\approx - 0.693\). Then, \(t=\frac{-\ln(2)}{-0.03}\). Substituting \(\ln(2)\approx0.693\), we have \(t=\frac{0.693}{0.03}=23.1\).

Answer:

\(23\)