Exponential-Decay Word Problems

What is the difference between the exponential-growth formula and the exponential-decay formula?

The only difference between exponential-growth equations and exponential-decay equations is that the constant for decay situations is negative. The equation itself is just the same as for exponential growth, but you should expect a negative value for the constant. If you get a positive value, you should probably go back and check your work.

Note that the particular variables used in the equation may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:

No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the decay constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.

To solve exponential decay word problems, you may be plugging one value into the exponential-decay equation, and solving for the required result. But you may need to solve two problems in one, where you use, say, the half-life information to find the decay constant (probably by solving the exponential equation by using logarithms), and then using that value to find whatever the exercise requested.

What is an example of a radioactive-decay exercise?

• Radio-isotopes of different elements have different half-lives. Magnesium-27 has a half-life of 9.45 minutes. What is the decay constant for Magnesium-27? Round to five decimal places.

The constant was negative, as expected, because this was a decay problem. If I'd ended up with a positive value, this would have signalled to me that I'd made a mistake somewhere.

(Note: Technically, there should be units on the constant, so that the units on the time variable t will cancel off. In other words, technically, the answer should be "k = −0.07335/minute. But this probably won't matter in your math class.)

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• Some people are frightened of certain medical tests because the tests involve the injection of radioactive materials. A hepatobiliary scan of my gallbladder involved an injection of 0.5 cc's (or about one-tenth of a teaspoon) of Technetium-99m, which has a half-life of almost exactly 6 hours. While undergoing the test, I heard the technician telling somebody on the phone that "in twenty-four hours, you'll be down to background radiation levels." Figure out just how much radioactive material remained from my gallbladder scan after twenty-four hours. Show your work below.

It can be very useful, sometimes in unexpected places, to know that 1 cc (that is, one cubic centimeter) is equal to 1 mL (that is, one milliliter). I would recommend that you add this bit of knowledge to your store.

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By the way: Technetium-99m is one of the most commonly used radioisotope for these medical purposes. Its radiation is extremely low-energy, so the chance of mutation is very low. (Whatever you're being treated for is the greater danger.) The half-life is just long enough for the doctors to have time to take their pictures. The dose I was given is about as large as these injections typically get. Your body does not easily absorb this chemical, so most of the injection is... "voided", shall we say? into the sewer system. 🚽

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What is an example of solving a carbon-dating exercise?

• Carbon-14 has a half-life of 5730 years. You are presented with a document which purports to contain the recollections of a Mycenaean soldier during the Trojan War. The city of Troy was finally destroyed in about 1250 BC, or about 3250 years ago. Carbon-dating evaluates the ratio of radioactive carbon-14 to stable carbon-12. Given the amount of carbon-12 contained a measured sample cut from the document, there would have been about 1.3 × 10⁻¹² grams of carbon-14 in the sample when the parchment was new, assuming the proposed age is correct. According to your equipment, there remains 1.0 × 10⁻¹² grams. Is there a possibility that this is a genuine document? Or is this instead a recent forgery? Justify your conclusions.