# When a living organism dies, the amount of Carbon-14 in its system begins to decrease exponentially. Given that the half-life of Carbon-14 is 5,730 years, answer the following: a.) What is the exact value of the rate of decay for Carbon-14?

## b.) If a bone is found to contain 61% of the Carbon-14 in an equivalent living organism, how old is the bone? (Round to the nearest year) The bone is approximately years old. c.) After 6250 years what percent of Carbon-14 will remain in the bone? (Round to the nearest tenth of a percent) Approximately % will remain after 6250 years.

##### 1 Answer
Jan 14, 2018

$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$
$4 , 086$ years
47.0%

#### Explanation:

$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{k t}$
where $t$ is in years

$y \left(5730\right) = a \cdot {\left(\frac{1}{2}\right)}^{1} = \frac{1}{2} a$

$y \left(5730\right) = a \cdot {\left(\frac{1}{2}\right)}^{\left(\frac{1}{5730}\right) \cdot 5730}$

this means that $k$ is $\frac{1}{5730}$.

$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{\left(\frac{1}{5730}\right) \cdot t}$

$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$

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$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$
where $t$ is in years

$y \left(t\right) = 0.61 a$

$a \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{5730}} = 0.61 a$

${\left(\frac{1}{2}\right)}^{\frac{t}{5730}} = 0.61$

$\frac{t}{5730} = {\log}_{\frac{1}{2}} 0.61$

$t = {\log}_{\frac{1}{2}} 0.61 \cdot 5730$

$t = 4 , 086$ (nearest $1$)

$y \left(t\right) = a \cdot {\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$

$y \left(6250\right) = a \cdot {\left(\frac{1}{2}\right)}^{\frac{6250}{5730}}$

${\left(\frac{1}{2}\right)}^{\frac{6250}{5730}} = 0.469517 \ldots$

0.469517 * 100% = 47.0% (nearest 0.1%)