Question

Please answer the following questions using exponential and logarithmic models.

4) A wooden artifact from an archaeological dig contains 70 percent of the Carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of Carbon-14 is 5730 years.) In years

5) A tumor is injected with 0.5 grams of Iodine-125, which has a decay rate of 1.15% per day. Write an exponential model representing the amount of Iodine-125 remaining in the tumor after t days. Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 60 day

6) A doctor prescribes 100 milligrams of a therapeutic drug that decays by about 20% each hour. To the nearest hour, what is the half-life of the drug?

7) A tumor is injected with 0.4 grams of Iodine-125, which has a decay rate of 1.15% per day. To the nearest day, how long will it take for half of the Iodine-125 to decay? How many Days?

8) A tumor is injected with 0.4 grams of Iodine-125, which has a decay rate of 1.15% per day. Write an exponential model representing the number of grams f of Iodine-125 remaining in the tumor after t days.

f(t)=?

Then use the formula to find the amount of Iodine-125 that would remain in the tumor after 30 days. (Round your answer to four decimal places.) In grams

9) The half-life of Erbium-171 is 7.5 hours. What is the hourly decay rate? Express the result to four decimal places.

Express the hourly decay rate as a percentage to two decimal places. In percent %

Answer 1

4. Let the initial amount of Carbon be N₀

Amount of carbon after t years = N₀ e^-kt

Given, half life = 5730 years

So, N₀/2 = N₀ e^-k(5730)

=> 1/2 = e^-5730k

=> ln (e^-5730k) = ln (1/2)

=> -5730k = ln(1/2)

=> k = 0.000121

Let the artifact be t years old

0.7N₀ = N₀ e^-0.000121t

=> 0.7 = e^-0.000121t

=> ln (e^-0.000121t) = ln 0.7

=> -0.000121t = ln (0.7)

=> t = 2947.73 years

5. 1-day decay percent = 1.15% = 0.0115

For x percent decrease , decay factor g is given by

g = 1 - 0.0115 = 0.9885

Initial amount = 0.5 grams

Amount of Iodine after t days = Initial Amount * (1-day decay factor)^t

=> A(t) = 0.5(0.9885)^t

Amount of Iodine after 60 days = 0.5(0.9885)⁶⁰ = 0.2498 gm

6. 1-hour decay percent = 20% = 0.2

  For x percent decrease , decay factor g is given by

g = 1 - 0.2 = 0.8

Let the initial amount be N_o

Amount of drug after t hours = N_o (0.8)^t

Let the half life be n hours

So, N_o/2 = N_o (0.8)ⁿ

=> (0.8)ⁿ = 0.5

=> ln (0.8)ⁿ = ln 0.5

=> n ln 0.8 = ln 0.5

=> n = 3 hours

So, the half life of the drug is 3 hours