Question

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).
$y = y_0 e^{-0.0936t}$
find the half - life of this substance. do not round any intermediate computations, and round your answer to the nearest tenth.
$square$ days

Explanation:

Step1: Use the half - life formula

The formula for radioactive decay is \(y = y_0e^{-kt}\). At half - life, \(y=\frac{y_0}{2}\). So, \(\frac{y_0}{2}=y_0e^{-0.0936t}\).

Step2: Simplify the equation

Divide both sides by \(y_0\) (since \(y_0
eq0\)), we get \(\frac{1}{2}=e^{- 0.0936t}\).

Step3: Take the natural logarithm of both sides

\(\ln(\frac{1}{2})=\ln(e^{-0.0936t})\). Using the property \(\ln(e^x)=x\) and \(\ln(\frac{1}{2})=-\ln(2)\), we have \(-\ln(2)=-0.0936t\).

Step4: Solve for \(t\)

\(t = \frac{\ln(2)}{0.0936}\). Since \(\ln(2)\approx0.6931\), then \(t=\frac{0.6931}{0.0936}\approx7.4\).

Answer:

\(7.4\)