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A scientist measures the initial amount of Carbon-14 in a substance to be 25 grams.
The relationship between 
A, the amount of Carbon-14 remaining in that substance, in grams, and 
t, the elapsed time, in years, since the initial measurement is modeled by the following equation.

A=25e^(-0.00012 t)
In how many years will the substance contain exactly 20 grams 
(g) of Carbon14 ?
Give an exact answer expressed as a natural logarithm.
years

A scientist measures the initial amount of Carbon14-14 in a substance to be 2525 grams.\newlineThe relationship between \newlineAA, the amount of Carbon14-14 remaining in that substance, in grams, and \newlinett, the elapsed time, in years, since the initial measurement is modeled by the following equation.\newlineA=25e0.00012tA=25e^{-0.00012 t}\newlineIn how many years will the substance contain exactly 2020 grams \newline(g) of Carbon1414 ?\newlineGive an exact answer expressed as a natural logarithm.\newlineyears

Full solution

Q. A scientist measures the initial amount of Carbon14-14 in a substance to be 2525 grams.\newlineThe relationship between \newlineAA, the amount of Carbon14-14 remaining in that substance, in grams, and \newlinett, the elapsed time, in years, since the initial measurement is modeled by the following equation.\newlineA=25e0.00012tA=25e^{-0.00012 t}\newlineIn how many years will the substance contain exactly 2020 grams \newline(g) of Carbon1414 ?\newlineGive an exact answer expressed as a natural logarithm.\newlineyears
  1. Write Given Exponential Decay Equation: Write down the given exponential decay equation.\newlineThe equation provided is A=25e(0.00012t)A = 25e^{(-0.00012t)}, where AA is the amount of Carbon14-14 remaining and tt is the time in years.
  2. Set A Equal to 2020 Grams: Set AA equal to 2020 grams to solve for tt. We want to find out when the substance will contain exactly 2020 grams of Carbon14-14, so we set AA to 2020. 20=25e0.00012t20 = 25e^{-0.00012t}
  3. Divide by 2525 to Isolate Exponential Term: Divide both sides of the equation by 2525 to isolate the exponential term.\newline(20/25)=e(0.00012t)(20/25) = e^{(-0.00012t)}\newline0.8=e(0.00012t)0.8 = e^{(-0.00012t)}
  4. Take Natural Logarithm to Solve for t: Take the natural logarithm of both sides to solve for t.\newlineln(0.8)=ln(e0.00012t)\ln(0.8) = \ln(e^{-0.00012t})
  5. Simplify Right Side of Equation: Use the property of logarithms that ln(ex)=x\ln(e^x) = x to simplify the right side of the equation.ln(0.8)=0.00012t\ln(0.8) = -0.00012t
  6. Divide by 0.00012-0.00012 to Solve for tt: Divide both sides by 0.00012-0.00012 to solve for tt.\newlinet=ln(0.8)0.00012t = \frac{\ln(0.8)}{-0.00012}
  7. Leave Answer as Natural Logarithm: Leave the answer as a natural logarithm as requested.\newlineThe final answer is expressed as a natural logarithm.

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