# intriguing question..involving logs and exponential decay/growth?

• Oct 24th 2009, 11:41 PM
BuffaloSoulja
intriguing question..involving logs and exponential decay/growth?
Carbon taken from an old animal skeleton contains 3/4 as much radioactive carbon14 as carbon taken from a present day bone. How old is the animal skeleton? (the half-life of carbon 14 is 5760 years)

halflife: the time required for something to fall to half its initial value

let P represent the amount of radioactive carbon 14 from a present day bone

i have 2 problems with this question.

i know 1/2P= P(x)^5760, but what adjustment will be made to 5760 since we are dealing with 3/4P instead of 1/2P?

and how do you isolate x after that?

• Oct 25th 2009, 01:21 AM
earboth
Quote:

Originally Posted by BuffaloSoulja
Carbon taken from an old animal skeleton contains 3/4 as much radioactive carbon14 as carbon taken from a present day bone. How old is the animal skeleton? (the half-life of carbon 14 is 5760 years)

halflife: the time required for something to fall to half its initial value

let P represent the amount of radioactive carbon 14 from a present day bone

i have 2 problems with this question.

i know [1] 1/2P= P(x)^5760, but what adjustment will be made to 5760 since we are dealing with 3/4P instead of 1/2P?

and how do you isolate x after that?

1. Calculate the base from [1]:

$\dfrac12 P = P \cdot x^{5760}$ Divide by P and calculate the 5760th root:

$x = \sqrt[5760]{\dfrac12} = \left(\dfrac12\right)^{\frac1{5760}} = 2^{-\frac1{5760}}$

2. According to the question you know:

$\dfrac34 P = P \cdot \left(2^{-\frac1{5760}} \right)^t$

Divide by P and solve the equation for t. Use the base-change-formula to calculate a logarithm:

$t = \log_{\left(2^{-\frac1{5760}}\right)}\left(\dfrac34 \right)$

Spoiler:
I've got $t \approx 2390\ years$