# Simple Logarithm

1. Dec 5, 2008

### icystrike

1. The problem statement, all variables and given/known data
It was found that the percentage of the carbon-14,P, contained in the bones of an animal n years after it has died is given by P = 2$$^{-kn}$$ , where k is a positive constant. The percentage of carbon-14 contained in the bones after the animal has been dead for 5668 years was 50% . How long, to the nearest year, was the animal dead if the percentage of the carbon-14 found in the bones was 76%?

I tried however my K is negative.

2. Relevant equations

3. The attempt at a solution

2. Dec 5, 2008

### HallsofIvy

Staff Emeritus
You tried? Good! Now, what did you try? Show exactly how you got K negative and we may be able to help you. (Perhaps you used P= 50 instead of 0.50?)

3. Dec 5, 2008

### icystrike

why is it 0.5 and not 50? i thought they were asking for the percentage and not the fraction?

The k i got is log , base 2, 50 over -5668

4. Dec 5, 2008

### Дьявол

You use 0.5 because 50%=50/100=5/10=1/2=2-1
$$2^{5668k}=2^{-1}$$
Now you need to find k and then substitute in 2-kn=76/100 and find n

5. Dec 5, 2008

### HallsofIvy

Staff Emeritus
Where did you get the idea that they were asking for a percentage? The problem specifically said " How long, to the nearest year". "How long" is NOT a percentage!