# Halflife question

1. Feb 16, 2010

### DanialD

1. The problem statement, all variables and given/known data

An archaeologist finds an oak wine cask in one of her digs. Testing the activity from the radioactive carbon-14 in the cask reveals that it is only one-quarter that of the activity coming from the modern sample of the same type of oak. How old is the sample?

2. Relevant equations
A=Anaught(.5)^(t/t1/2)

3. The attempt at a solution
I tried finding the halflife of carbon-14 (which is 5.73EXP3a) from a chart given in my textbook.
after that, i tried filling int he blanks for my equation above, but i just didnt know how to formulate it.

2. Feb 16, 2010

### collinsmark

You could use your equation to solve this one. The amount of carbon 14 based radioactivity is proportional to the amount of Carbon 14 present.

$$A = A _0 (0.5)^{\frac{t}{t_{1/2}}}$$

Here, $$A_0$$ can be considered the activity of the modern sample. Since it is being used as the reference of comparison, you can set it equal to 1. Then $$A$$ becomes the activity of a sample relative to a modern sample (0.25 in this case). $$t _{1/2}$$ is the halflife. Then just solve for $$t$$.

There's also an easier way to do this particular problem. If the activity is 1/2 of that compared to a modern sample, then the age is one halflife. If the activity is 1/2 of 1/2, then the age is two halflives, and so on.

3. Feb 16, 2010

### DanialD

Thank you so much, you helped me a lot.
Just a question, is there another way to formulate this question using log?

4. Feb 16, 2010

### collinsmark

There certainly is! Just take the logarithm of both sides of the equation, and note that

$$Log(x^a) = aLog(x)$$

or more specifically in this case,

$$Log \left( \left(0.5 \right) ^\frac{t}{t_{1/2}} \right) = \frac{t}{t_{1/2}}Log(0.5)$$

You'd have to do that anyway when solving for t. So in other words, take the logarithm of both sides (noting the above identity), and solve for t.

Btw, I used the common logarithm, but you can use the natural logarithm if you'd like or logarithms of any base for that matter. Whatever you prefer.

5. Feb 16, 2010

### DanialD

again, thanks so much!

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