To find the age of the fossil bone, we start with the formula for the amount of carbon-14 remaining in the object:
\[ y = a e^{-0.00012t} \]
Here, \(y\) is the remaining amount of carbon-14, \(a\) is the original amount, and \(t\) is the age of the object in years.
In this problem, we know that the fossil bone contains 24% of its original carbon-14. This means:
\[ y = 0.24a \]
Now we can substitute \(y\) into the equation:
\[ 0.24a = a e^{-0.00012t} \]
We can divide both sides by \(a\) (assuming \(a \neq 0\)):
\[ 0.24 = e^{-0.00012t} \]
Next, we take the natural logarithm of both sides to solve for \(t\):
\[ \ln(0.24) = -0.00012t \]
Now, solving for \(t\):
\[ t = \frac{\ln(0.24)}{-0.00012} \]
Calculating \(\ln(0.24)\):
\[ \ln(0.24) \approx -1.4271 \]
Now substitute this value into the equation for \(t\):
\[ t = \frac{-1.4271}{-0.00012} \approx 11892.5 \]
Therefore, the approximate age of the bone is:
\[ t \approx 11893 \text{ years} \]
So, the age of the fossil bone is approximately 11,893 years.