Math / Algebra

QuestionA wooden artifact from an archeological dig contains 73 percent of the carbon-14 that is present in living trees. To the nearest year, about how many years old is the artifact? (The half-life of carbon14 is 5730 years.)
The artifact is $\square$ years old. Enter an integer or decimal number [more.] Question Help: Written Example Submit Question

Studdy Solution

STEP 1

What is this asking? How many years have passed since the artifact had 100% of its carbon-14, given it now has 73% and carbon-14's half-life is 5730 years? Watch out! Don't mix up the current and initial amounts of carbon-14!
Also, remember the half-life is the time it takes for the substance to reduce to half its initial amount, not to zero!

STEP 2

1. Set up the exponential decay formula
2. Solve for the time elapsed

STEP 3

We'll use the exponential decay formula: $N(t) = N_0 \cdot e^{-\lambda t}$ .
Here, $N(t)$ is the amount of carbon-14 remaining after time $t$ , $N_0$ is the initial amount of carbon-14, $\lambda$ is the decay constant, and $t$ is the time in years we want to find.

STEP 4

We know that the artifact currently has 73% of its initial carbon-14.
So, $N(t) = 0.73 N_0$ .
Let's plug that into our formula: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .

STEP 5

We're given the half-life of carbon-14, which is 5730 years.
This means that when $t = 5730$ , $N(t) = \frac{1}{2} N_0$ .
Plugging this into the formula, we get $\frac{1}{2} N_0 = N_0 \cdot e^{-\lambda \cdot 5730}$ .
Dividing both sides by $N_0$ gives us $\frac{1}{2} = e^{-\lambda \cdot 5730}$ .

STEP 6

To solve for $\lambda$ , we'll take the natural logarithm (ln) of both sides: $\ln(\frac{1}{2}) = \ln(e^{-\lambda \cdot 5730})$ .
This simplifies to $\ln(\frac{1}{2}) = -\lambda \cdot 5730$ .
Now, we can solve for $\lambda$ : $\lambda = -\frac{\ln(\frac{1}{2})}{5730} = \frac{\ln(2)}{5730} \approx \mathbf{0.000121}$ .

STEP 7

Now we have everything we need to solve for $t$ !
Remember our equation from earlier: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .
Divide both sides by $N_0$ to get $0.73 = e^{-\lambda t}$ .

STEP 8

Just like we did before, let's take the natural logarithm of both sides: $\ln(0.73) = \ln(e^{-\lambda t})$ .
This simplifies to $\ln(0.73) = -\lambda t$ .

STEP 9

Now, plug in the value of $\lambda$ we found earlier: $\ln(0.73) = -0.000121 \cdot t$ .
Finally, solve for $t$ : $t = -\frac{\ln(0.73)}{0.000121} \approx \mathbf{2622}$ .

STEP 10

The artifact is approximately 2622 years old.

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Studdy Solution

STEP 1

STEP 2

1. Set up the exponential decay formula
2. Solve for the time elapsed

STEP 3

We'll use the exponential decay formula: $N(t) = N_0 \cdot e^{-\lambda t}$ .
Here, $N(t)$ is the amount of carbon-14 remaining after time $t$ , $N_0$ is the initial amount of carbon-14, $\lambda$ is the decay constant, and $t$ is the time in years we want to find.

STEP 4

We know that the artifact currently has 73% of its initial carbon-14.
So, $N(t) = 0.73 N_0$ .
Let's plug that into our formula: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .

STEP 5

We're given the half-life of carbon-14, which is 5730 years.
This means that when $t = 5730$ , $N(t) = \frac{1}{2} N_0$ .
Plugging this into the formula, we get $\frac{1}{2} N_0 = N_0 \cdot e^{-\lambda \cdot 5730}$ .
Dividing both sides by $N_0$ gives us $\frac{1}{2} = e^{-\lambda \cdot 5730}$ .

STEP 6

STEP 7

Now we have everything we need to solve for $t$ !
Remember our equation from earlier: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .
Divide both sides by $N_0$ to get $0.73 = e^{-\lambda t}$ .

STEP 8

Just like we did before, let's take the natural logarithm of both sides: $\ln(0.73) = \ln(e^{-\lambda t})$ .
This simplifies to $\ln(0.73) = -\lambda t$ .

STEP 9

Now, plug in the value of $\lambda$ we found earlier: $\ln(0.73) = -0.000121 \cdot t$ .
Finally, solve for $t$ : $t = -\frac{\ln(0.73)}{0.000121} \approx \mathbf{2622}$ .

STEP 10

The artifact is approximately 2622 years old.

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Studdy Solution

STEP 1

STEP 2

1. Set up the exponential decay formula2. Solve for the time elapsed

STEP 3

STEP 4

We know that the artifact currently has 73% of its initial carbon-14.So, N(t)=0.73N0N(t) = 0.73 N_0N(t)=0.73N0​.Let's plug that into our formula: 0.73N0=N0⋅e−λt0.73 N_0 = N_0 \cdot e^{-\lambda t}0.73N0​=N0​⋅e−λt.

STEP 5

STEP 6

STEP 7

Now we have everything we need to solve for ttt!Remember our equation from earlier: 0.73N0=N0⋅e−λt0.73 N_0 = N_0 \cdot e^{-\lambda t}0.73N0​=N0​⋅e−λt.Divide both sides by N0N_0N0​ to get 0.73=e−λt0.73 = e^{-\lambda t}0.73=e−λt.

STEP 8

Just like we did before, let's take the natural logarithm of both sides: ln⁡(0.73)=ln⁡(e−λt)\ln(0.73) = \ln(e^{-\lambda t})ln(0.73)=ln(e−λt).This simplifies to ln⁡(0.73)=−λt\ln(0.73) = -\lambda tln(0.73)=−λt.

STEP 9

Now, plug in the value of λ\lambdaλ we found earlier: ln⁡(0.73)=−0.000121⋅t\ln(0.73) = -0.000121 \cdot tln(0.73)=−0.000121⋅t.Finally, solve for ttt: t=−ln⁡(0.73)0.000121≈2622t = -\frac{\ln(0.73)}{0.000121} \approx \mathbf{2622}t=−0.000121ln(0.73)​≈2622.

STEP 10

The artifact is approximately **2622** years old.

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1. Set up the exponential decay formula
2. Solve for the time elapsed

We know that the artifact currently has 73% of its initial carbon-14.
So, $N(t) = 0.73 N_0$ .
Let's plug that into our formula: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .

Now we have everything we need to solve for $t$ !
Remember our equation from earlier: $0.73 N_0 = N_0 \cdot e^{-\lambda t}$ .
Divide both sides by $N_0$ to get $0.73 = e^{-\lambda t}$ .

Just like we did before, let's take the natural logarithm of both sides: $\ln(0.73) = \ln(e^{-\lambda t})$ .
This simplifies to $\ln(0.73) = -\lambda t$ .

Now, plug in the value of $\lambda$ we found earlier: $\ln(0.73) = -0.000121 \cdot t$ .
Finally, solve for $t$ : $t = -\frac{\ln(0.73)}{0.000121} \approx \mathbf{2622}$ .

The artifact is approximately 2622 years old.