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Math

Math Snap

PROBLEM

The amount of carbon-14 present in animal bones after tt years is given by A(t)=A0e0.00012tA(t) = A_0 e^{-0.00012t}. A sample of fossil had 28% of the carbon 14 of a contemporary living sample. Estimate the age of the sample.
The age of the sample is ___ years.
(Round to the nearest year as needed.)

STEP 1

What is this asking?
We're figuring out how old a fossil is based on how much carbon-14 is left, knowing that carbon-14 decays over time.
Watch out!
Don't mix up the initial amount with the current amount.
Also, make sure to round to the nearest year at the very end!

STEP 2

1. Set up the equation
2. Isolate the exponential
3. Solve for tt

STEP 3

The problem says the fossil has \textbf{28%} of the carbon-14 of a living sample.
This means A(t)A(t) is 0.280.28 times the initial amount, A0A_0.
So we can write A(t)=0.28A0A(t) = 0.28 \cdot A_0.

STEP 4

We know the formula for carbon-14 decay is A(t)=A0e0.00012tA(t) = A_0 e^{-0.00012t}.
Let's substitute the expression for A(t)A(t) we just found:
0.28A0=A0e0.00012t0.28 \cdot A_0 = A_0 e^{-0.00012t}

STEP 5

To get the exponential term by itself, we can divide both sides of the equation by A0A_0.
This is valid since A0A_0 represents the initial amount of carbon-14, which is a positive value.
0.28A0A0=A0e0.00012tA0\frac{0.28 \cdot A_0}{A_0} = \frac{A_0 e^{-0.00012t}}{A_0} 0.28=e0.00012t0.28 = e^{-0.00012t}

STEP 6

To get tt out of the exponent, we can take the natural logarithm (ln) of both sides:
ln(0.28)=ln(e0.00012t)\ln(0.28) = \ln(e^{-0.00012t})

STEP 7

Remember that ln(ex)=x\ln(e^x) = x, so we have:
ln(0.28)=0.00012t\ln(0.28) = -0.00012t

STEP 8

Now, we can divide both sides by 0.00012-0.00012 to solve for tt:
t=ln(0.28)0.00012t = \frac{\ln(0.28)}{-0.00012}

STEP 9

Using a calculator, we find:
t1.2729656750.00012t \approx \frac{-1.272965675}{-0.00012} t10608.04729t \approx 10608.04729

STEP 10

The problem asks us to round to the nearest year, so our final answer is approximately 10608\textbf{10608} years.

SOLUTION

The age of the sample is approximately 10608 years.

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