Finding half-life or doubling time A radioactive substance decays according to the following function, where $y_{0}$ is the initial amount present, and $y$ is the amount present at time $t$ (in days). $y=y_{0} e^{-0.0741 t}$ Find the half-life of this substance. Do not round any intermediate computations, and round your answer to the nearest tenth. П్) days $\times 5$

Step 1 ：We are given a radioactive substance that decays according to the function \(y=y_{0} e^{-0.0741 t}\), where \(y_{0}\) is the initial amount present, \(y\) is the amount present at time \(t\) (in days), and \(-0.0741\) is the decay rate.

Step 2 ：We need to find the half-life of this substance, which is the time it takes for the substance to reduce to half its initial amount. In other words, we need to find the time \(t\) when \(y = \frac{y_0}{2}\).

Step 3 ：We can set up the equation \(\frac{y_0}{2} = y_{0} e^{-0.0741 t}\) and solve for \(t\).

Step 4 ：Dividing both sides of the equation by \(y_0\), we get \(\frac{1}{2} = e^{-0.0741 t}\).

Step 5 ：Taking the natural logarithm of both sides, we get \(\ln(\frac{1}{2}) = -0.0741 t\).

Step 6 ：Solving for \(t\), we get \(t = \frac{\ln(\frac{1}{2})}{-0.0741}\).

Step 7 ：Calculating the right side of the equation, we find that the half-life of the substance is approximately 9.4 days.

Step 8 ：Final Answer: The half-life of the substance is \(\boxed{9.4}\) days.