atuck please help

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 a less radioactive and more stable form. The function for the half-life of an isotope is shown below.

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

1. Find the inverse of the function .A(t) = 500(1/2)^t/272 Make sure you enter each step and you can't use natural log

2 . What does the input variable of the inverse function represent? (5 points)

3. What does the output variable of the inverse function represent? (5 points)

Answer 1

1)

A/A₀ = (1/2)^t/h

log₂(A/A₀)=(t/h)log₂(1/2) = (t/h)(-1) since log₂(1/2) is -1

t(A)=-hlog₂(A/A₀) = -272log₂(A/500) Note: A/500 must be < 1

2) "A" is the current amount of the isotope

3) t(A) is the amount of time to get to the 1/2 life given A