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 8000 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.

The decay model is given by   W(x) = 2^(–x/24360) where x is the number of years and W(x) is the amount (in kilograms) remaining from an initial 1 kilogram. (a) For x = 5000 years, substitute into the function:   W(5000) = 2^(–5000/24360) Calculate the exponent:   5000/24360 ≈ 0.2054 Thus:   W(5000) ≈ 2^(–0.2054) To convert to an exponential function:   2^(–0.2054) = exp(–0.2054 × ln 2) Since ln 2 ≈ 0.6931,   W(5000) ≈ exp(–0.2054 × 0.6931) = exp(–0.1423) ≈ 0.8677 So, approximately 0.87 kilograms remain after 5000 years. (b) For x = 8000 years:   W(8000) = 2^(–8000/24360) Calculate the exponent:   8000/24360 ≈ 0.3284 Thus:   W(8000) ≈ 2^(–0.3284) = exp(–0.3284 × ln 2)       ≈ exp(–0.3284 × 0.6931) = exp(–0.2274) ≈ 0.7966 So, approximately 0.80 kilograms remain after 8000 years. (c) For x = 16,000 years:   W(16,000) = 2^(–16000/24360) Calculate the exponent:   16000/24360 ≈ 0.6568 Thus:   W(16,000) ≈ 2^(–0.6568) = exp(–0.6568 × ln 2)       ≈ exp(–0.6568 × 0.6931) = exp(–0.455) ≈ 0.6346 So, approximately 0.63 kilograms remain after 16,000 years. (d) To find the time it takes for 1 kilogram to decay to half its original weight, we set W(x) equal to 0.5:   W(x) = 2^(–x/24360) = 0.5 But 0.5 can be written as 2^(–1), so we have:   2^(–x/24360) = 2^(–1) Since the bases are the same, the exponents must be equal:   –x/24360 = –1 Multiply both sides by –24360:   x = 24360 Thus, it takes 24,360 years for the plutonium to decay to half its original weight. Summary:  (a) After 5000 years: approximately 0.87 kg remain.  (b) After 8000 years: approximately 0.80 kg remain.  (c) After 16,000 years: approximately 0.63 kg remain.  (d) The half-life is 24,360 years.