Solve Logarithm: Find n if P=76%, k=-5668, P=2^{-kn}

[icystrike]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

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?)

 

[icystrike]

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?)

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

 

[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

 

[HallsofIvy]

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

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!