Find inverse, etc

Question

Cobalt -57 is used in medical testing and has a half-life of about 272 days, which means after 272 days, half of the Cobalt -57 has decayed to less radioactive and more stable form. The function for the half-life of an isotope is shown.

A(t)=A₀(1/2)^t/h

A₀ is the initial amount of the isotope, h is the half-life, t is the input or time elapse, and A(t) is the output or the amount of the isotope remaining as a function of time.

1)Find the inverse of A(t)=500*(1/2)^t/500 and list the steps. CANNOT use natural log (LN).

2)What does the input variable of the inverse represent?

3)What does the output variable of the inverse function represent?

4)If 100 grams of the cobalt-57 has decayed to 62 grams, how much time has passed?

Answer 1

2. The input variable is the amount of Cobalt-57 remaining at the end of the time period.

3. The output variable is the number of days that the Cobalt-57 has been decaying.

Answer 2

Since the half-life is 272, the exponent should be t/272, not t/500.

A = A(t) = 500(1/2)^t/272, where t = number of days, A = amount remaining after t days, and 500 = initial amount

A/500 = (1/2)^t/272

A/500 = 2^-t/272

-t / 272 = log₂(A / 500) = log₂A - log₂500

t = -272(log₂A - log₂500)

If initial amount = 100 and A = 62, then t = -272(log₂62 - log₂100) = 187.6 days