# Suppose a bone from an ancient human contains 16.5% of the C-14 found in living organisms. How old is the bone? Please help; thank you so much!!?

Mar 31, 2016

The bone is approximately $14894.92$ years old.

#### Explanation:

When dealing with a half life question, it is best to use the half-life formula, which is expressed as:

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} y = a {\left(b\right)}^{\frac{t}{h}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where:
$y =$final amount
$a =$inital amount
$b =$growth/decay
$t =$time elapsed
$h =$half-life

Note that the half-life of carbon-14 is approximately $5730$ years.

$1$. Start by expressing 16.5% as a decimal and plugging the value into the half-life formula as well as your other known values.

• $a = 1$ is 100%, expressed as a decimal
• $b = \frac{1}{2}$ indicates half-life

$y = a {\left(b\right)}^{\frac{t}{h}}$

$0.165 = 1 {\left(\frac{1}{2}\right)}^{\frac{t}{5730}}$

$2$. Since the bases are not the same on both sides of the equation, take the logarithm of both sides.

$\log \left(0.165\right) = \log \left({\left(\frac{1}{2}\right)}^{\frac{t}{5730}}\right)$

$3$. Use the logarithmic property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to simplify the right side of the equation.

$\log \left(0.165\right) = \left(\frac{t}{5730}\right) \log \left(\frac{1}{2}\right)$

$4$. Isolate for $t$.

$\frac{t}{5730} = \frac{\log \left(0.165\right)}{\log \left(\frac{1}{2}\right)}$

$t = = \frac{5730 \log \left(0.165\right)}{\log \left(\frac{1}{2}\right)}$

$5$. Solve.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} t \approx 14894.92 \textcolor{w h i t e}{i} \text{years old} \textcolor{w h i t e}{\frac{a}{a}} |}}}$