A hiker in Africa discovers a skull that contains 63% of its original amount of C-14. N=Noe^-kt No=initial amount of C-14(at time t=0) N=amount of C-14 at time t t=time, in years k=0.0001

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A hiker in Africa discovers a skull that contains 63% of its original amount of C-14. N=Noe^-kt No=initial amount of C-14(at time t=0) N=amount of C-14 at time t t=time, in years k=0.0001

Mathematics
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Ok so we have: \[ N(t)=N_o e^{-kt}\]
We are told that their is 63% of C14 remaining in the material hence: \[ \frac{N(t)}{N_o}= 63\% =0.63= e^{-kt}\]
yes but it says find the age of the skull to the nearest year.

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Taking the ln of both sides: \[\ln(0.63)= -(0.0001 yr^{-1})t\]
Divide and you have you answer
divide 0.63 / -0.0001?
Please note it isnt 0.63 but ln(0.63)
And then it is a yes
Btw, in case your worried about the minus sign... the fact that the argument of the logarithm is <1 means its result will be <0.... so the minus signs cancel and it will yield a positive value for t
i got -6300
natrual log of 0.63 divided by -10^-4
Please note that the CRUCIAL step in this problem was taking the log of both sides. Without that step you cannot reduce the exponential factor.
your confusing me
What do you find confusing?
Please, I cant help if I dont know where your getting confused.
Go through the process I did at the top and let me know which step you are uneasy about and I will give a more detailed explanation.
show him the steps in detail
I was under the impression I did alekos but I will go through it again.
Ok so we given in the problem that the amount of C14 we find in the skull is 63%
This means that of the original amount: \[N_0=100\% (some \ amount \ of \ nuclei)=(some \ amount \ of \ nuclei) \] It decays according to the law \[N(t)=N_0 e^{-kt}\] Where the N's correspond to the number of nuclei of the material in quesion remaining (N(t)) or the original amount (N_0) So that at time T, we have: \[N(T)=63\% (some \ amount \ of \ nuclei)=(some \ amount \ of \ nuclei) e^{-kT}\] Now whatever the total amount of nucleai there are in the sample is irrelevant because I can just divide both sides by that amount leaving just the percent remaining.
This results in the equation I gave above: \[63\% = 0.63=e^{-kT}\]
Now just like when we have an equation of the form: \[2=x^2\] Which we solve by applying the inverse function of squared (i.e. the square root) to both sides (ignoring the +/- that comes in for simplicity): \[\sqrt{2}=\sqrt{x^2}=x\] When we have an exponential equation of the form: \[2=e^x\] We apply the inverse function to the exponential function (aka the natural logarithm) to both sides in order to simplfy and solve: \[ \ln(2)=\ln(e^x)=x\]
Note that an exponential function can NEVER be negative (unless I explicitly multiply it by a -1), which means when I do this inverse function business applying the logarithm, I dont have to worry about a +/- like you do when you apply a square root. So don't let that trouble you.
So performing this step (taking the natural log of both sides of the equation) yields: \[\ln{0.63}=\ln{e^{-kT}}=-kT=-(10^{-4}yr^{-1})T\] Then solve for T, by dividing both sides by k: \[T=\frac{\ln{0.63}}{-(10^{-4}yr^{-1})}\]
Please calculate this value and express it with the appropriate units here I and I will check your answer.
@ix.ty are you still here?
im on a diffrent question now
A country's population in 1992 was 222 million. In 2001 it was 224 million. Estimate the population in 2004 using the exponential growth formula. Round your answer to the nearest million. P = Aekt
So what did you get for the last one?
i got it wrong that was my last question but i passed the test
Oh so before I even try to start what looks to be a HARDER question on the same material it bears going over the last problem to iron out what confused you. Please tell me what you found confusing.
No judgement, I fancy myself like a bit of an auto-mechanic.... I need to know where the problem is in order to try and fix it.
dont bother plasma, seems like you're wasting you're time
Perhaps, but I am stubborn and banging my head against the wall might be a bit of a pastime for me.... But either way I am willing to help @ix.ty but I will only stick around for maybe 10-20 minutes (Ill check back at this tab) then I will move on

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