Math / Calculus

QuestionSuppose an Egyptian mummy is discovered in which the amount of carbon-14 present is only about three-fourths the amount found in living human beings. The amount of carbon-14 present in animal bones after $t$ years is given by $y=y_{0} e^{-0.0001216 t}$ , where $y_{0}$ is the amount of carbon-14 present in living human beings. About how long ago did the Egyptian die?
About $\square$ yr ago the Egyptian had died. (Round to the nearest integer as needed.)

Studdy Solution

STEP 1

What is this asking? We're trying to figure out how many years ago an Egyptian died based on how much carbon-14 is left in their remains compared to a living person. Watch out! Don't forget to solve for $t$ , and make sure to use the natural logarithm correctly!

STEP 2

1. Set up the equation
2. Solve for $t$
3. Round the answer

STEP 3

Alright, let's dive in!
We know that the amount of carbon-14 left in the mummy is three-fourths of what's found in living humans.
So, we can set up the equation using the formula given:
$y = y_{0} e^{-0.0001216 t}$
Here, $y = \frac{3}{4}y_{0}$ because the mummy has three-fourths of the carbon-14.
Plug that into our equation:
$\frac{3}{4}y_{0} = y_{0} e^{-0.0001216 t}$

STEP 4

Now, let's solve for $t$ !
First, we can divide both sides by $y_{0}$ to simplify:
$\frac{3}{4} = e^{-0.0001216 t}$

STEP 5

Next, we need to take the natural logarithm of both sides to get rid of that pesky $e$ :
$\ln\left(\frac{3}{4}\right) = \ln\left(e^{-0.0001216 t}\right)$

STEP 6

Using the property of logarithms that says $\ln(e^x) = x$ , we can simplify the right side:
$\ln\left(\frac{3}{4}\right) = -0.0001216 t$

STEP 7

Finally, let's solve for $t$ by dividing both sides by $-0.0001216$ :
$t = \frac{\ln\left(\frac{3}{4}\right)}{-0.0001216}$

STEP 8

Now, let's calculate that using a calculator:
$t \approx \frac{\ln(0.75)}{-0.0001216} \approx 2401.2$
Since we need to round to the nearest integer, we get:
$t \approx 2401$

STEP 9

About 2401 years ago, the Egyptian had died.

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Studdy Solution

STEP 1

What is this asking? We're trying to figure out how many years ago an Egyptian died based on how much carbon-14 is left in their remains compared to a living person. Watch out! Don't forget to solve for $t$ , and make sure to use the natural logarithm correctly!

STEP 2

1. Set up the equation
2. Solve for $t$
3. Round the answer

STEP 3

STEP 4

Now, let's solve for $t$ !
First, we can divide both sides by $y_{0}$ to simplify:
$\frac{3}{4} = e^{-0.0001216 t}$

STEP 5

Next, we need to take the natural logarithm of both sides to get rid of that pesky $e$ :
$\ln\left(\frac{3}{4}\right) = \ln\left(e^{-0.0001216 t}\right)$

STEP 6

Using the property of logarithms that says $\ln(e^x) = x$ , we can simplify the right side:
$\ln\left(\frac{3}{4}\right) = -0.0001216 t$

STEP 7

Finally, let's solve for $t$ by dividing both sides by $-0.0001216$ :
$t = \frac{\ln\left(\frac{3}{4}\right)}{-0.0001216}$

STEP 8

Now, let's calculate that using a calculator:
$t \approx \frac{\ln(0.75)}{-0.0001216} \approx 2401.2$
Since we need to round to the nearest integer, we get:
$t \approx 2401$

STEP 9

About 2401 years ago, the Egyptian had died.

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Studdy Solution

STEP 1

What is this asking? We're trying to figure out how many years ago an Egyptian died based on how much carbon-14 is left in their remains compared to a living person. Watch out! Don't forget to solve for t t t, and make sure to use the natural logarithm correctly!

STEP 2

1. Set up the equation2. Solve for t t t3. Round the answer

STEP 3

STEP 4

Now, let's solve for t t t!First, we can **divide both sides by y0 y_{0} y0​** to simplify:34=e−0.0001216t\frac{3}{4} = e^{-0.0001216 t}43​=e−0.0001216t

STEP 5

Next, we need to **take the natural logarithm** of both sides to get rid of that pesky e e e:ln⁡(34)=ln⁡(e−0.0001216t)\ln\left(\frac{3}{4}\right) = \ln\left(e^{-0.0001216 t}\right)ln(43​)=ln(e−0.0001216t)

STEP 6

Using the property of logarithms that says ln⁡(ex)=x\ln(e^x) = xln(ex)=x, we can simplify the right side:ln⁡(34)=−0.0001216t\ln\left(\frac{3}{4}\right) = -0.0001216 tln(43​)=−0.0001216t

STEP 7

Finally, let's **solve for t t t** by dividing both sides by −0.0001216-0.0001216−0.0001216:t=ln⁡(34)−0.0001216t = \frac{\ln\left(\frac{3}{4}\right)}{-0.0001216}t=−0.0001216ln(43​)​

STEP 8

Now, let's calculate that using a calculator:t≈ln⁡(0.75)−0.0001216≈2401.2t \approx \frac{\ln(0.75)}{-0.0001216} \approx 2401.2t≈−0.0001216ln(0.75)​≈2401.2Since we need to round to the nearest integer, we get:t≈2401t \approx 2401t≈2401

STEP 9

About **2401** years ago, the Egyptian had died.

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Studdy solves anything!

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What is this asking? We're trying to figure out how many years ago an Egyptian died based on how much carbon-14 is left in their remains compared to a living person. Watch out! Don't forget to solve for $t$ , and make sure to use the natural logarithm correctly!

1. Set up the equation
2. Solve for $t$
3. Round the answer

Now, let's solve for $t$ !
First, we can divide both sides by $y_{0}$ to simplify:
$\frac{3}{4} = e^{-0.0001216 t}$

Next, we need to take the natural logarithm of both sides to get rid of that pesky $e$ :
$\ln\left(\frac{3}{4}\right) = \ln\left(e^{-0.0001216 t}\right)$

Using the property of logarithms that says $\ln(e^x) = x$ , we can simplify the right side:
$\ln\left(\frac{3}{4}\right) = -0.0001216 t$

Finally, let's solve for $t$ by dividing both sides by $-0.0001216$ :
$t = \frac{\ln\left(\frac{3}{4}\right)}{-0.0001216}$

Now, let's calculate that using a calculator:
$t \approx \frac{\ln(0.75)}{-0.0001216} \approx 2401.2$
Since we need to round to the nearest integer, we get:
$t \approx 2401$

About 2401 years ago, the Egyptian had died.