Questions: Homework 2.5: Exponential and Logarithmic Models Score: 15 / 100 Answered: 2 / 10 Question 3 At the beginning of an experiment, a scientist has 396 grams of radioactive goo. After 240 minutes, her sample has decayed to 49.5 grams. What is the half-life of the goo in minutes? Find a formula for G(t), the amount of goo remaining at time t. G(t)= How many grams of goo will remain after 26 minutes? You may enter the exact value or round to 2 decimal places.

Solution Steps

To find the half-life of the radioactive goo, we can use the formula for exponential decay:

$A(t) = A_0 \cdot e^{-kt}$

where:

• $A(t)$ is the amount of substance remaining at time $t$
• $A_0$ is the initial amount of substance
• $k$ is the decay constant
• $t$ is the time elapsed

We can use the given information to set up an equation to solve for the half-life.

To find the amount of goo remaining at time $t$, we can use the formula $G(t) = 396 \cdot e^{-kt}$.

To find the half-life, we can set up the equation $49.5 = 396 \cdot e^{-240k}$ and solve for $k$.

To find the amount of goo remaining after 26 minutes, we can use the formula $G(26) = 396 \cdot e^{-26k}$.

Given:

• Initial amount \( A_0 = 396 \)
• Final amount \( A(t) = 49.5 \)
• Time elapsed for half-life \( t = 240 \)

Using the formula \( k = -\frac{\ln(A(t) / A_0)}{t} \), we find: \[ k = -\frac{\ln(49.5 / 396)}{240} \approx 0.0086643 \]

Given:

• Time elapsed \( t = 26 \)

Using the formula \( G(t) = 396 \cdot e^{-kt} \), we find: \[ G(26) = 396 \cdot e^{-0.0086643 \cdot 26} \approx 316.13 \]

Final Answer