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June 2017, Part IV

The Question in Part IV is worth up to 6 credits. Partial credit is possible.


37. A radioactive substance has a mass of 140 g at 3 p.m. and 100 g at 8 p.m. Write an equation in the form A = A₀(1/2)^(t/h) that models this situation, where h is the constant representing the number of hours in the half-life, A₀ is the initial mass, and A is the mass t hours after 3 p.m.

Using this equation, solve for h, to the nearest ten thousandth.

Determine when the mass of the radioactive substance will be 40 g. Round your answer to the nearest tenth of an hour.

Answer:
We are given A = 100, A₀ = 140, and t = (8 - 3) = 5. We want h.
A = A₀(1/2)^(t/h)
100 = 140 (1/2)^(5/h)
100/140 = (1/2)^(5/h)
5 / 7 = (1/2)^(5/h)
(log 5/7) / (log 1/2) = (5/h) log (1/2) / log (1/2)
(log 5/7) / (log 1/2) = 5 / h
(log 5/7) * h = (log 1/2) * 5
h = (log 1/2) / (log 5/7) * 5
h = 10.3002, to the nearest ten-thousandth.

Second part:
We are given A = 40, A₀ = 140. We now have h = 10.3002. We want t.
A = A₀(1/2)^(t/h)
40 = 140 (1/2)^(t/10.3002)
40 / 140 = (1/2)^(t/10.3002)
2 / 7 = (1/2)^(t/10.3002)
(log 2 / 7) / (log 1/2) = t / 10.3002
t = 10.3002 * (log 2/7) / (log 1/2)
t = 18.61... = 18.6 hours to the nearest tenth of an hour.

Logs were never my favorite subject in school, even if the rules are pretty straightforward. It's just one more inverse operation.

Next up: January 2017.



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