solve his using S=V(1-r)^t instead of the half life rule---A radioactive substance decays according to the law A(t) = A_0 (1/2)^{t/T_{1/2}} where T_{1/2} is the half-life. (a) If the initial amount is A_0 = 200 g and the half-life is T_{1/2} = 5 years, find the time t (in years) required for the sample to decay to 50 g. (b) A piece of machinery depreciates in value continuously at rate k per year according to V(t) = V_0 e^{-kt}. If V_0 = 10,000 dollars and its value falls to 6,000 dollars in 4 years, determine the depreciation rate k (to three decimal places).

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Let's solve this radioactive decay problem. We have a substance with initial amount 200 grams and half-life of 5 years. We need to find when it decays to 50 grams. The decay follows the exponential formula A of t equals A naught times one-half to the power of t over T half. Looking at the graph, we can see the exponential decay curve starting at 200 grams. Now let's solve this step by step. We substitute our known values into the decay equation: 50 equals 200 times one-half to the power of t over 5. Dividing both sides by 200 gives us one-fourth equals one-half to the power of t over 5. Since one-fourth is the same as one-half squared, we can equate the exponents: 2 equals t over 5. Solving for t gives us t equals 10 years. Notice how the substance halves every 5 years: from 200 to 100 grams in the first half-life, then from 100 to 50 grams in the second half-life. Now let's move to part b, the machinery depreciation problem. We have a machine worth 10,000 dollars initially that depreciates continuously according to V of t equals V naught times e to the negative k t. After 4 years, its value drops to 6,000 dollars. We need to find the depreciation rate k. This is a continuous exponential decay model, different from the discrete half-life model we just solved. Let's solve for the depreciation rate k step by step. We substitute our known values: 6000 equals 10000 times e to the negative 4k. Dividing both sides by 10000 gives us 0.6 equals e to the negative 4k. To solve for k, we take the natural logarithm of both sides. Using the property that ln of e to the x equals x, we get ln of 0.6 equals negative 4k. Finally, solving for k gives us k equals ln of 0.6 divided by negative 4, which equals approximately 0.128. This means the machinery depreciates at a rate of 12.8 percent per year. Let's summarize our solutions. For part a, the radioactive decay problem, we found that it takes 10 years for the sample to decay from 200 grams to 50 grams. For part b, the machinery depreciation problem, we calculated the depreciation rate k to be approximately 0.128 per year. The key difference between these problems is that radioactive decay follows a discrete half-life model, while machinery depreciation uses a continuous exponential decay model. Both are exponential processes, but they use different mathematical formulations to describe the decay behavior.