Question

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

Answer 1

Ok so we have:
\[ N(t)=N_o e^{-kt}\]

Answer 2

We are told that their is 63% of C14 remaining in the material hence:
\[ \frac{N(t)}{N_o}= 63\% =0.63= e^{-kt}\]

Answer 3

yes but it says find the age of the skull to the nearest year.

Answer 4

Taking the ln of both sides:
\[\ln(0.63)= -(0.0001 yr^{-1})t\]

Answer 5

Divide and you have you answer

Answer 6

divide 0.63 / -0.0001?

Answer 7

Please note it isnt 0.63 but ln(0.63)

Answer 8

And then it is a yes

Answer 9

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

Answer 10

i got -6300

Answer 11

natrual log of 0.63 divided by -10^-4

Answer 12

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.

Answer 13

your confusing me

Answer 14

What do you find confusing?

Answer 15

Please, I cant help if I dont know where your getting confused.

Answer 16

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.

Answer 17

show him the steps in detail

Answer 18

I was under the impression I did alekos but I will go through it again.

Answer 19

Ok so we given in the problem that the amount of C14 we find in the skull is 63%

Answer 20

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.

Answer 21

This results in the equation I gave above:
\[63\% = 0.63=e^{-kT}\]

Answer 22

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:
\[ \ln(2)=\ln(e^x)=x\]

Answer 23

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.

Answer 24

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:
\[T=\frac{\ln{0.63}}{-(10^{-4}yr^{-1})}\]

Answer 25

Please calculate this value and express it with the appropriate units here I and I will check your answer.

Answer 26

@ix.ty are you still here?

Answer 27

im on a diffrent question now

Answer 28

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

Answer 29

So what did you get for the last one?

Answer 30

i got it wrong that was my last question but i passed the test

Answer 31

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.

Answer 32

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.

Answer 33

dont bother plasma, seems like you're wasting you're time

Answer 34

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