Questions: A radioactive substance decays according to the following function, where y0 is the initial amount present, and y is the amount present at time t (in days). y = y0 e^(-0.0162 t) Find the half-life of this substance. Do not round any intermediate computations, and round your answer to the nearest tenth. days

Solution Steps

We are given a decay function for a radioactive substance: $$y = y_{0} e^{-0.0162 t}$$ We need to find the half-life of the substance, which is the time $t$ when the amount $y$ is half of the initial amount $y_{0}$.

The half-life $t_{1/2}$ is the time at which $y = \frac{y_{0}}{2}$. Substitute $y = \frac{y_{0}}{2}$ into the decay equation: $$\frac{y_{0}}{2} = y_{0} e^{-0.0162 t_{1/2}}$$

Divide both sides by $y_{0}$: $$\frac{1}{2} = e^{-0.0162 t_{1/2}}$$

Take the natural logarithm of both sides to solve for $t_{1/2}$: $$\ln\left(\frac{1}{2}\right) = \ln\left(e^{-0.0162 t_{1/2}}\right)$$ $$\ln\left(\frac{1}{2}\right) = -0.0162 t_{1/2}$$ $$t_{1/2} = \frac{\ln\left(\frac{1}{2}\right)}{-0.0162}$$

Calculate the natural logarithm and divide: $$t_{1/2} = \frac{\ln(0.5)}{-0.0162} \approx \frac{-0.6931}{-0.0162} \approx 42.7778$$

Round the computed half-life to the nearest tenth: $$t_{1/2} \approx 42.8$$

Final Answer