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Suppose an Egyptian mummy is discovered in which the amount of carbon-14 present is only about two-fifths the amount found in living human beings. The amount of carbon-14 present in animal bones after [tex][tex]$t$[/tex][/tex] years is given by [tex][tex]$y=y_0 e^{-0.0001216 t}$[/tex][/tex], where [tex][tex]$y_0$[/tex][/tex] is the amount of carbon-14 present in living human beings. About how long ago did the Egyptian die?

About [tex]\square[/tex] years ago the Egyptian had died.
(Round to the nearest integer as needed.)

Sagot :

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.
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