Respuesta :
Answer:
7705.43 years.
Step-by-step explanation:
Let us assume that initial amount of mastodon was 1.
We have been given that the radioactive element carbon-14 has a half-life of 5750 years. A scientist determined that the bones from a mastodon had lost 60.5% of their carbon-14.
We will half life formula to solve our given problem.
[tex]A=a\cdot (\frac{1}{2})^{\frac{t}{h}}[/tex], where,
A= Final amount left after t years,
a = Initial amount,
t = time,
h = Half-life of substance.
Since the bones from a mastodon had lost 60.5% of their carbon-14, so amount of carbon-14 left would be:
[tex]1-60.5\%\rightarrow 1-\frac{60.5}{100}=1-0.605=0.395[/tex]
Upon substituting our given values in half'life formula, we will get:
[tex]0.395=1\cdot (\frac{1}{2})^{\frac{t}{5750}}[/tex]
[tex]0.395=(0.5)^{\frac{t}{5750}}[/tex]
Upon taking natural log of both sides, we will get:
[tex]\text{ln}(0.395)=\text{ln}((0.5)^{\frac{t}{5750}})[/tex]
Using log property [tex]\text{ln}(a^b)=b\cdot \text{ln}(a)[/tex], we will get:
[tex]\text{ln}(0.395)=\frac{t}{5750}\times\text{ln}(0.5)[/tex]
[tex]\text{ln}(0.395)=\frac{t\times\text{ln}(0.5)}{5750}[/tex]
Now, we will divide both sides of our equation by [tex]\text{ln}(0.5)[/tex].
[tex]\frac{\text{ln}(0.395)}{\text{ln}(0.5)}=\frac{t\times\text{ln}(0.5)}{5750\times \text{ln}(0.5)}}[/tex]
[tex]\frac{\text{ln}(0.395)}{\text{ln}(0.5)}=\frac{t}{5750}[/tex]
Switch sides:
[tex]\frac{t}{5750}=\frac{\text{ln}(0.395)}{\text{ln}(0.5)}[/tex]
[tex]\frac{t}{5750}=\frac{-0.9288695}{-0.693147180559}[/tex]
[tex]\frac{t}{5750}=1.34007544159762[/tex]
[tex]\frac{t}{5750}\times 5750=1.34007544159762\times 5750[/tex]
[tex]t=7705.433789186[/tex]
[tex]t\approx 7705.43[/tex]
Therefore, the bones were approximately 7705.43 years old when they were discovered.
