# The half-life of carbon-14 is 5700 years. What is the age to the nearest year of a sample in which 39% of the radioactive nuclei originally present have decayed?

Mar 26, 2016

The sample is approximately $4065$ 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

$1$. Start by substituting your known values into the formula.

• In this case, y=100%-39%=61%, since 39% of the sample decayed, leaving 61% to remain. Express $y$ as $61$, derived from 61%, but without the "%."
• Express $a$ in the equation as $100$, derived from 100%, but without the "%."
• The value of b=100%-"decay rate", or in your case, b=100%-50%=50% or expressed as a fraction, $\frac{1}{2}$.

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

$61 = 100 {\left(\frac{1}{2}\right)}^{\frac{t}{5700}}$

$2$. Divide both sides by $100$.

$\frac{61}{100} = {\left(\frac{1}{2}\right)}^{\frac{t}{5700}}$

$3$. Since the bases on both sides of the equation are not the same, take the logarithm of both sides.

$\log \left(\frac{61}{100}\right) = \log \left({\left(\frac{1}{2}\right)}^{\frac{t}{5700}}\right)$

$4$. Use the log 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 rewrite the right side of the equation.

$\log \left(\frac{61}{100}\right) = \frac{t}{5700} \cdot \log \left(\frac{1}{2}\right)$

$5$. Isolate for $t$.

$\frac{t}{5700} = \frac{\log \left(\frac{61}{100}\right)}{\log \left(\frac{1}{2}\right)}$

$t = \frac{5700 \cdot \log \left(\frac{61}{100}\right)}{\log \left(\frac{1}{2}\right)}$

$6$. Solve for $t$.

$t = 4064.78 \ldots$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} t \approx 4065 \textcolor{w h i t e}{i} \text{years old} \textcolor{w h i t e}{\frac{a}{a}} |}}}$