donnie1999
  • donnie1999
Situation: An archaeologist in turkey discovers a spear head that contains 32% of its original amount of c-14. N= Noe^-kt No= initial amount of C-14 at time t t= time, in years k= 0.0001 Find the age of the spear head to the nearest year.
Algebra
chestercat
  • chestercat
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donnie1999
  • donnie1999
donnie1999
  • donnie1999
anonymous
  • anonymous

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phi
  • phi
start with \[ N= N_0e^{-kt} \] N is how much you have now, N0 is how much you started with. if we divide both sides by N0, you have \[ \frac{N}{N_0} = e^{-kt} \] N/N0 is the fraction we now have left. They tell us this is 32%, which means 32/100 or 0.32 also, they tell us k= 0.0001 put those numbers into the equation \[ 0.32= e^{-0.0001\cdot t} \]
phi
  • phi
I would "take the natural log" of both sides as the next step.
donnie1999
  • donnie1999
whats that?
phi
  • phi
If you are doing this type of problem, they assume you know about logarithms
donnie1999
  • donnie1999
@phi I am kind of clueless to these types of questions I was hoping you could help me out a little more.
phi
  • phi
natural log , written ln means "exponent of base e" for example, if you write \[ e^x \] the natural log of that \[ \ln\left(e^x\right) = x \] is the exponent
phi
  • phi
you want to use that idea on your equation to "get" the exponent -0.0001*t so write ln ( e^-0.0001*t) but because it is an equation, do the same to the other side, and write ln(0.32) now simplify the right-hand side using the rule: ln( e^stuff) simplifies to stuff

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