Question

the decay equation for a certain substance is known to be $y = y_0e^{-0.0330t}$, with $t$ in days. about how long will it take the substance in a sealed sample of air to fall to 60% of its original value? it will take approximately days. (round to one decimal place as needed.)

Explanation:

Step1: Set up the equation

We want the substance to fall to 60% of its original value. So $y = 0.6y_0$. Substitute into the decay - equation $y=y_0e^{-0.0330t}$. We get $0.6y_0=y_0e^{-0.0330t}$. Since $y_0
eq0$, we can divide both sides by $y_0$ and the equation becomes $0.6 = e^{-0.0330t}$.

Step2: Take the natural logarithm of both sides

Taking the natural logarithm of both sides of the equation $0.6 = e^{-0.0330t}$, we use the property $\ln(e^x)=x$. So $\ln(0.6)=\ln(e^{-0.0330t})$, which simplifies to $\ln(0.6)=- 0.0330t$.

Step3: Solve for t

We know that $\ln(0.6)\approx - 0.5108$. Then $t=\frac{\ln(0.6)}{-0.0330}$. Substituting the value of $\ln(0.6)$ into the formula, we have $t=\frac{-0.5108}{-0.0330}\approx15.5$.

Answer:

$15.5$