Questions: Exponential and Logarithmic Functions Finding half-life or doubling time Actinium-227 is a radioactive substance that decays according to the following function, where y₀ is the initial amount present, and y is the amount present at time t (in years). y=y₀ e^(-0.0321 t) Find the half-life of Actinium-227. Do not round any intermediate computations, and round your answer to the nearest tenth. years

Solution Steps

The decay function for Actinium-227 is given by: \[ y = y_{0} e^{-0.0321 t} \] where \( y_{0} \) is the initial amount and \( y \) is the amount at time \( t \) in years.

The half-life of a substance is the time it takes for half of the initial amount to decay. Mathematically, this means: \[ \frac{y_{0}}{2} = y_{0} e^{-0.0321 t_{1/2}} \] where \( t_{1/2} \) is the half-life.

To find the half-life, we need to solve for \( t_{1/2} \). Start by dividing both sides of the equation by \( y_{0} \): \[ \frac{1}{2} = e^{-0.0321 t_{1/2}} \]

Next, take the natural logarithm of both sides to solve for \( t_{1/2} \): \[ \ln\left(\frac{1}{2}\right) = \ln\left(e^{-0.0321 t_{1/2}}\right) \]

Since \( \ln(e^x) = x \), this simplifies to: \[ \ln\left(\frac{1}{2}\right) = -0.0321 t_{1/2} \]

Now, solve for \( t_{1/2} \): \[ t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-0.0321} \]

Using the property \( \ln\left(\frac{1}{2}\right) = -\ln(2) \): \[ t_{1/2} = \frac{-\ln(2)}{-0.0321} = \frac{\ln(2)}{0.0321} \]

Calculate the value: \[ \ln(2) \approx 0.6931 \] \[ t_{1/2} = \frac{0.6931}{0.0321} \approx 21.5888 \]

Final Answer