Need help with Alg Question ASAP thank you!

Question

The original function for this question is

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

Half-life is an interesting concept, it describes the time it takes for half of any given element to decay. We use the decay of carbon atoms to tell how old fossils are, as carbon decays at a constant rate. Each substance has its own unique half-life, or time it takes for half of the substance to decay into something else.

I am assuming the half-life of cobalt-57 was given in your problem, though I don't see it in your question so I will answer as best as I can. Google says the half-life of cobalt-57 is 271.79 days.

Let's define our variables:

A(o) = initial amount of cobalt-57 (grams)

A(t) = amount of cobalt-57 left after given time (grams)

t = time elapsed

h = half-life of cobalt, 271.79 days

1. Find the inverse of A(t)=A(o)(1/2)^t/h

To find the inverse, we will switch our input (t or x) and output (A(t) or y) variables and solve for y.

Replace A(t) with y and t with x to keep things more simple. A(o) and h are constants so they should remain as they are.

The equation becomes:

y= A(o)(1/2)^x/h

Now we switch x and y values;

x=A(o)(1/2)^y/h

Now we solve for y:

divide both sides by A(o)

x/A(o) = (1/2)^y/h

given that log_ba=c is equivalent to b^c=a;

log_(1/2)(x/A_o) = y/h

multiply both sides of the equation by h;

h*[log_(1/2)(x/A_(o))] = y

now we have our inverse equation:

h*[log_(1/2)(x/A_(o))] = y

(2) Now that we have our inverse function, what do the variables represent?

Let's try a simpler example. Let's say we have the equation y=2x, where y is the amount of cookies eaten, and x is the amount of time passed eating cookies. Each x value represents an amount of time eating, and each y value represents the amount of cookies eaten in that time. If we invert the equation, we can start with any given amount of cookies, and find how much time it took to eat them.

The inverse, x/2 = y, would tell us how much time (y) it took to eat the given (x) number of cookies.

So, for half-life, the inverse of the equation, rather than telling us how much cobalt-57 is left after a given amount of time, we can use a given amount of remaining cobalt-57 to calculate the time that has passed.

The input, x, is the remaining cobalt-57 in grams.

(3) The output, y, is the time passed in the degradation reaction.

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

We can solve this problem using either the given or inverse equation. I will use the inverse because we have designed it to output a time based on a given amount of cobalt-57.

h(log_1/2(x/A(o))=y

let's define our variables:

h = half-life of cobalt-57: 271.79 days

x = amount of cobalt left in reaction: 62 grams

A_o = initial amount of cobalt-57 : 100 grams

y = time passed in reaction

so;

271.79(log_1/2(62/100)) = y

log_1/2(0.62) = 0.68965987939

(271.79)(0.68965987939) = 187.442658619

y=~ 187.4 days