help needed asap!

Question

The original function for this question is A(t) = A₀(1/2)^t/h .

This question is about cobalt-57 and half life.

1. Find the inverse of the function from question 2. You can't use natural log (LN).

2 . What does the input variable of the inverse function represent?

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

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

Answer 1

The formula for exponential decay:P(the current amount)=PO ( the original amount(100) e^(-kt). K=270 for cobolt 79 t=time or

62=100 e^(-270t) T=185 days

Cobalt (Co) 57 Isotope Radioactivity Decay Calculation

Isotope (t)

Cobalt 57

Half-life (T1/2)

270

days

Initial Activity (A0)

100

Decay time (t)

185

days

Final Activity (A)

62.198871261220326

I did not adhere to your teacher question.

I googled decay calculator and retrieved the above.

The deay formula is DIFFERENT from other formulas and one has to be careful!

The formula for isotope decay i AP=AO x e^(-kt)

AP=present amount

AO=original amount

k=half life

t=time elapsed

One can use a different base than the natural logs but it is foolish and leads to a rather complicated formula.

ezcalc.meEasy calculate online!

1. Home
2. Math Calculators
3. Exponential Decay Calculator

Exponential Decay Calculator

This all-in-one online Exponential Decay Calculator evaluates the continuous exponential decay function. It can be also used as Half Life Calculator. You can enter the values of any three parameters in the input fields of this calculator and find the missing parameter.

Initial Amount: 

Decay Rate: 

Time Passed: 

Final Amount: 

Exponential decay formula

The exponential decay process can be expressed by the following formula:

A(t)=A(0)e−rt

where A(t) and A(0) are amounts of some quantity at time t and 0 respectively, r is the decay rate and t is the time passed.

The exponential decay is found in processes where amount of something decreases at a rate proportional to its current value. Exponential decay occurs in a wide variety of cases that mostly fall into the domain of the natural sciences. The most famous example is radioactive decay. But this phenomenon can also be found in chemical reactions, pharmacology and toxicology, physical optics, electrostatics, luminescence and many more.

I hope that helps!

Answer 2

Let me first an answer questions 2) and 3)

In the original function there are two input variables:

t, representing time, and

A0 representing initial amount of cobalt.

There is one output variable A, representing the amount of cobalt left after time t.

In the inverse function the roles of variables are reversed.

The variable A will definitely be an input variable.

But question 1. is ambiguous -- we have two choices for output variable: t and A0.

Choosing t to be an input variable, and A0 an output variable would allow is to

solve problems of the form

"What was the original amount A0 of cobalt if we observed a given amount A left

after given time t?"

Choosing A0 to be an input variable, and t an output variable would allow is to

solve problems of the form

"How long did it take for a given amount A0 of cobalt to decay to the given amount A?

Question 4. disambiguates question 1. -- your teacher wants you to solve

the latter kind of problems.

That is, t is the output variable and A0 is a second input variable.

Now we can answer question 1 converting the original function to one

expressing how to calculate t form A and A0.

Maybe by now you know to to do it. But just to make sure, let me do it step by step.

And I will follow your teacher's suggestion to use the natural log

(as oppose to log base 2).

A = A0 (1/2)^t/h

A = A0 (e^ln(1/2))^t/h

ln(A) = ln(A0 (e^ln(1/2))^t/h)

ln(A) = ln(A0) + t/h ln(1/2)

ln(A) = ln(A0) - t/h ln(2)

t = h(ln(A0) - ln(A))/ln(2)

t = h ln(A0/A) / ln(2)

The above expression is then the answer to question 1.

For question 4

h = 271.79 days

A0 = 100 g

A = 62 g

t = 271.79 ln(100/62) / ln(2) = 187.44 days