Answer :
To determine how long ago the Egyptian died, we can use the given exponential decay formula for carbon-14 in the mummy:
[tex]\[ y = y_0 e^{-0.0001216 t} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial amount of carbon-14 in living human beings.
- [tex]\( y \)[/tex] is the amount of carbon-14 found in the mummy.
- [tex]\( t \)[/tex] is the time in years since the Egyptian died.
- The decay constant is [tex]\( 0.0001216 \)[/tex].
We know that the amount of carbon-14 found in the mummy is about two-fifths of the amount found in living beings, thus:
[tex]\[ y = \frac{2}{5} y_0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ \frac{2}{5} y_0 = y_0 e^{-0.0001216 t} \][/tex]
Next, we can simplify by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ \frac{2}{5} = e^{-0.0001216 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{2}{5}\right) = \ln\left(e^{-0.0001216 t}\right) \][/tex]
Since the natural logarithm and the exponential function are inverse functions, we can simplify further:
[tex]\[ \ln\left(\frac{2}{5}\right) = -0.0001216 t \][/tex]
Now, solve for [tex]\( t \)[/tex] by isolating it on one side:
[tex]\[ t = \frac{\ln\left(\frac{2}{5}\right)}{-0.0001216} \][/tex]
Evaluating this expression, we find:
[tex]\[ t \approx 7535.28562396509 \][/tex]
Rounding to the nearest integer gives us:
[tex]\[ t \approx 7535 \][/tex]
Therefore, the mummy had died approximately 7535 years ago.
About [tex]\( \boxed{7535} \)[/tex] years ago the Egyptian had died.
[tex]\[ y = y_0 e^{-0.0001216 t} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial amount of carbon-14 in living human beings.
- [tex]\( y \)[/tex] is the amount of carbon-14 found in the mummy.
- [tex]\( t \)[/tex] is the time in years since the Egyptian died.
- The decay constant is [tex]\( 0.0001216 \)[/tex].
We know that the amount of carbon-14 found in the mummy is about two-fifths of the amount found in living beings, thus:
[tex]\[ y = \frac{2}{5} y_0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ \frac{2}{5} y_0 = y_0 e^{-0.0001216 t} \][/tex]
Next, we can simplify by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ \frac{2}{5} = e^{-0.0001216 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{2}{5}\right) = \ln\left(e^{-0.0001216 t}\right) \][/tex]
Since the natural logarithm and the exponential function are inverse functions, we can simplify further:
[tex]\[ \ln\left(\frac{2}{5}\right) = -0.0001216 t \][/tex]
Now, solve for [tex]\( t \)[/tex] by isolating it on one side:
[tex]\[ t = \frac{\ln\left(\frac{2}{5}\right)}{-0.0001216} \][/tex]
Evaluating this expression, we find:
[tex]\[ t \approx 7535.28562396509 \][/tex]
Rounding to the nearest integer gives us:
[tex]\[ t \approx 7535 \][/tex]
Therefore, the mummy had died approximately 7535 years ago.
About [tex]\( \boxed{7535} \)[/tex] years ago the Egyptian had died.
