Question: The amount of plutonium remaining from 1 kilogram after \( x \) years is given by the function…

The amount of plutonium remaining from 1 kilogram after \( x \) years is given by the function \( W(x) = 2^{-\frac{x}{24,360}} \).

(a) How much will be left after 5000 years?

(b) How much will be left after 11,000 years?

(c) How much will be left after 16,000 years?

(d) Estimate how long it will take for the 1 kilogram to decay to half its original weight.

Solution

The function for the amount of plutonium remaining is given by: \[ W(x) = 2^{-\frac{x}{24,360}} \] (a) To find the amount left after 5000 years: Substitute \( x = 5000 \) into the function: \[ W(5000) = 2^{-\frac{5000}{24,360}} \] Calculate the exponent: \[ W(5000) = 2^{-0.2053} \] Calculate \( W(5000) \): \[ W(5000) \approx 0.8684 \] (b) To find the amount left after 11,000 years: Substitute \( x = 11,000 \) into the function: \[ W(11,000) = 2^{-\frac{11,000}{24,360}} \] Calculate the exponent: \[ W(11,000) = 2^{-0.4514} \] Calculate \( W(11,000) \): \[ W(11,000) \approx 0.7358 \] (c) To find the amount left after 16,000 years: Substitute \( x = 16,000 \) into the function: \[ W(16,000) = 2^{-\frac{16,000}{24,360}} \] Calculate the exponent: \[ W(16,000) = 2^{-0.6571} \] Calculate \( W(16,000) \): \[ W(16,000) \approx 0.6445 \] (d) To estimate the time to decay to half its original weight: Set \( W(x) = 0.5 \) and solve for \( x \): \[ 0.5 = 2^{-\frac{x}{24,360}} \] Take the logarithm of both sides: \[ \log(0.5) = -\frac{x}{24,360} \log(2) \] Solve for \( x \): \[ x = \frac{\log(0.5)}{-\log(2)} \times 24,360 \] Calculate \( x \): \[ x \approx 24,360 \] So, it takes approximately 24,360 years for the plutonium to decay to half its original weight.