Question

a certain radioactive isotope decays according to the model $y(t) = y_0 e^{-0.0002t}$, where $y(t)$ represents the amount of the isotope left after $t$ years.
question
in approximately how many years will there be 92% of the isotope left?
select a single answer
251 years
800 years
417 years
400 years

Explanation:

Step1: Set up decay equation

$0.92y_0 = y_0e^{-0.0002t}$

Step2: Cancel $y_0$ (non-zero)

$0.92 = e^{-0.0002t}$

Step3: Take natural log of both sides

$\ln(0.92) = -0.0002t$

Step4: Solve for t

$t = \frac{\ln(0.92)}{-0.0002}$

Step5: Calculate value

$t \approx \frac{-0.08338}{-0.0002} = 416.9 \approx 417$
Wait, correction: Recalculating $\ln(0.92)$: $\ln(0.92) \approx -0.08338$, so $t = \frac{-0.08338}{-0.0002} = 416.9 \approx 417$ years.

Step1: Set up decay equation

$0.92y_0 = y_0e^{-0.0002t}$

Step2: Cancel $y_0$

$0.92 = e^{-0.0002t}$

Step3: Apply natural logarithm

$\ln(0.92) = -0.0002t$

Step4: Isolate t

$t = \frac{\ln(0.92)}{-0.0002}$

Step5: Compute numerical value

$t \approx \frac{-0.08338}{-0.0002} \approx 417$

Answer:

251 years