Half Life: Find a, R for geometric sequence for iodine-131 decay

[joshdal9]

How do you solve this using number patterns (preferably not logs)?

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8. Nuclear medicine procedures help detect and treat diseases by using a small amount of radioactive material called a radiopharmaceutical. Iodine-131 is used to diagnose and treat thyroid cancer and has a half-life (the time take for 50% of the iodine-131 to decay) of 8 days. A patient is given a single dose of 5mg of iodine-131 in their system.

a. Identify the first term, a, and the common ratio, R, for the geometric sequence for the radioactive decay.
b. Calculate the amount of iodine-131 remaining after 24 days.

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This is my working so far (may or may not be correct):
5mg
Half life - 8 days
24 days
A = 5(1/2)^t/8
=5(1/2)^24/8
=5(1/2)^3
A=5/8
Part A - A = 5 R =1/2

 

[stapel]

How do you solve this using number patterns (preferably not logs)?

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8. Nuclear medicine procedures help detect and treat diseases by using a small amount of radioactive material called a radiopharmaceutical. Iodine-131 is used to diagnose and treat thyroid cancer and has a half-life (the time take for 50% of the iodine-131 to decay) of 8 days. A patient is given a single dose of 5mg of iodine-131 in their system.

a. Identify the first term, a, and the common ratio, R, for the geometric sequence for the radioactive decay.
b. Calculate the amount of iodine-131 remaining after 24 days.

---

This is my working so far (may or may not be correct):
5mg
Half life - 8 days
24 days
A = 5(1/2)^t/8
=5(1/2)^24/8
=5(1/2)^3
A=5/8
Part A - A = 5 R =1/2

You seem to have two different values for "A" (which I'm guessing is meant to stand for "a", which is a different variable); you have stated that A = 5/8 and that A = 5. Which is it? Also, you seem to be using "minus" signs for things that you're not actually subtracting...? This can be confusing (and can annoy the grader, and what's the point of that, right?).

I *think* you mean something along the lines of the following:

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initial dose: 5mg
half-life: 8 days

a. The initial amount is the constant "a", so: a = 5
The half-life is 8 days so, if the geometric-sequence terms give the amount at the end of each consecutive eight-day period, then the amount will always be half of the preceding amount. Then the common ratio is R = 1/2.

b. Twenty-four days is three eight-day periods, so the sequence would be: 5, 5/2, 5/4, 5/8. So the amount remaining at the end of twenty-eight days is 5/8 of a gram.

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Is this correct? Thank you!

Eliz.