Question

the age of a rock is determined by the function, ( a(t)=a_{0}e^{-0.000002t} ), where isotope q, with an initial value of ( a_{0} ), decays at a constant rate 0.000002 per year, for a number of years (also known as the age of the rock), ( t ).
a sample originally has 80 grams of isotope q, decays, and now has 20 grams of the isotope. determine the current age of the sample, in years. round your answer to the nearest thousand years. make sure to include the correct place value commas in your answer.
the age of the sample ( approx ) years.

Explanation:

Step1: Substitute values into formula

Given \(A(t) = 20\), \(A_0=80\), and the formula \(A(t)=A_0e^{- 0.000002t}\). Substitute: \(20 = 80e^{-0.000002t}\).

Step2: Simplify the equation

Divide both sides by \(80\): \(\frac{20}{80}=e^{-0.000002t}\), so \(\frac{1}{4}=e^{-0.000002t}\).

Step3: Take the natural logarithm of both sides

\(\ln(\frac{1}{4})=\ln(e^{-0.000002t})\). Since \(\ln(e^{x}) = x\), we get \(\ln(\frac{1}{4})=- 0.000002t\).

Step4: Calculate \(\ln(\frac{1}{4})\)

\(\ln(\frac{1}{4})=\ln(1)-\ln(4)=0 - \ln(4)\approx - 1.386294\).

Step5: Solve for \(t\)

We have \(-1.386294=-0.000002t\). Then \(t=\frac{-1.386294}{-0.000002}\).
\(t = 693147\approx693000\) (rounded to the nearest thousand).

Answer:

\(693,000\)