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

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

- anonymous

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

Divide and you have you answer

- anonymous

divide 0.63 / -0.0001?

- anonymous

Please note it isnt 0.63 but ln(0.63)

- anonymous

And then it is a yes

- anonymous

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

- anonymous

i got -6300

- anonymous

natrual log of 0.63 divided by -10^-4

- anonymous

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.

- anonymous

your confusing me

- anonymous

What do you find confusing?

- anonymous

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

- anonymous

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.

- alekos

show him the steps in detail

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

@ix.ty are you still here?

- anonymous

im on a diffrent question now

- anonymous

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

- anonymous

So what did you get for the last one?

- anonymous

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

- anonymous

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.

- anonymous

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.

- alekos

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

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.