Camel Bone

How do you use a half life equation?

My teacher gave us formulas, but she didn’t tell us how to use them. She said if you just divide by two for the problems, it wouldn’t work every time (do you know why this is – I like to know).

A sample problem is:
A camel skeleton is found in the La Brea Tar Pits. A section of the tibia (a lower leg bone) is tested and found to have only 1/8th of the original 14C isotope. How long ago did the camel fall into the tar pit? Assume a half life of 6000 years.

Equations that I know but don’t know how to use:
A = Ao e^kt
A(t) = Ao e^kt
Which one of these two? Also, is kt negative?

Y = A(1/2)^th
Is this one okay to use? And could t be over h (t/h).

Thanks for the help!

Hi:

you can use anyone of the equation to solve this – However I don’t think you wrote them correctly {check to see if they have a ( -) for the ( t/h)

but I like

A = Ao e^-kt

So I’m going to use this one to solve your problem :

Step 1 :
Given

h = 6000 years
A = Ao e^-kt
A= 1/8 of Ao or .125

Ao = 1 * 10^23 atoms of the camel skeleton ( You can use any amount of atoms for Ao as long as you calculate with 1/8 of it is for A )
and 1 /8 of 1 * 10^23 atoms is 1.25 * 10^22
e = 2.718281828459

We are solving for t

so

A = Ao e^-kt – original equations

1.25e+22 = 1 * 10^23 * 2.718281828459^- (t/ 6000) – substituting A,Ao, e and h

1.25e+22/ 1 * 10^23 = 1 * 10^23 * 2.718281828459^- (t/ 6000) * 1/ 1 * 10^23 – Multiplying the reciprocal of a number to both sides of the equation to move it to the other side of the equation

1.25e+22/ 1 * 10^23 = 2.718281828459^- (t/ 6000) – Multiplication

.125 = 2.718281828459^ -(t/ 6000) – Division

ln( .125) = ln( 2.718281828459^ -(t/ 6000)) – using logarithm on both sides of the equation to remove the exponent

ln(.125) = – t/6000 ln( 2.718281828459) the power rule of logarthms

since ln ( 2.718281828459) = 1 – the anti log of exponent with the same base rule

ln(.125) = – t / 6000 – the multiplication of one identity

- 6000 ln (.125) = – t /6000* -6000 – Multiplying the multiplicative inverse of a number to both sides of the equation to move it to the other side of it

- 6000 ln (.125) = t – multiplication

-6000 * -2.079 441 541 680 = t – sloving the ln of .125

t = 12476.649250079016 – multiplication ( I rounded it to 12 decimal place here )

t = 12,476.649250079016 years ago

Proof or check

A = Ao e^-kt – original equations

1.25*10^22 = 1 * 10^23 * 2.718281828459^- (12,476.649250079016 / 6000) – substituting A,Ao,e,k,t with 1.25 *10^22, 1 *10^23, 2.718281828459, 12,476.649250079016 , and 6000

1.25 * 10^22 = 1* 10^23 * 2.718281828459 ^- ( 2.079 441 541 680) – dividing 12,476.649250079016 by 6000

1.25 * 10^22 = 1 * 10^23 ( .125) – solving the natural log of – 2.079 441 541 680

1.25 * 10^22 = 1.25 * 10^ 22 – multiplication of 1 * 10^23 and .125

It checks and equals

I hope this helps

for more info :

http://en.wikipedia.org/wiki/Half-life

http://www.askmehelpdesk.com/math-sciences/half-life-equation-365047.html

http://en.wikipedia.org/wiki/Exponential_decay

http://mathworld.wolfram.com/ExponentialDecay.html

http://www.askmehelpdesk.com/math-sciences/half-life-equation-365047.html

I hope this helps

4. Camel Bones and 5. Vexated