I need the answers to these questions plz

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.

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

1) To fine the inverse of this function we are sloving for ‘t’.

A(t)=500*(1/2)^t/500 is our function, thought I believe t/500 should be t/272 as 272 is the half-life.

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

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

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

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

ln(A(t))- ln(500)=(t/500)ln(1/2)

ln(A(t)) - ln(500)= (t/500)(-ln(2))

(In(A(t)) - ln(500))/(-ln(2))= (t/500)

t = -500(In(A(t)) - ln(500))/(ln(2))

To fully understand I would go over your inverse rules, for yourself so you don’t get lost. Such as the ln(1/2) -> -ln(2) is a rule where you can take the reciprocal though you have a negative to that value now.

2.The input variable A(t) in the function still represents the amount of the Colbat -57 that is left after testing.

3.The output variable t is still the amount of time it takes for an initial amount of colbat to reach decay to the remaining amount left.

4.Replace the values in the equation using the general equation t= -h(ln(A(t)-ln(Ao))/ln(2) such that you get ~188 days til 100 grams of the material decays to days.

That should be it. Other tutors feel free correct I missed something but I am confident this is it. Hope this helps.