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Diagram showing a scientist measuring Carbon-14 levels in a substance with the equation A = 25e^0.00012t. A thought bubble indicates the scientist calculating the time required for 20 grams of Carbon-14, with the answer expressed as ln(20/25)/0.00012.

Calculating Carbon-14: Reaching 20 Grams with Natural Logarithms

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Question: 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^0.00012t In how many years will the substance contain exactly 20 grams of Carbon-14 ? Give an exact answer expressed as a natural logarithm.

Asked by hayden94

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From the given equation, we know that: A = 25e^0.00012t We need to find the time t when the amount A is 20 grams. We do this by setting A = 20 and then solving for t. 20 = 25e^0.00012t First, divide both sides by 25: 20/25 = e^0.00012t 0.8 = e^0.00012t Since e^x = y is equivalent to x = ln(y), we can write above equation as: 0.00012t = ln(0.8) Finally, dividing by 0.00012 on both sides gives the value of t: t = ln(0.8) / 0.00012 Therefore, the substance will contain exactly 20 grams of Carbon-14 after ln(0.8) / 0.00012 years.

Answered by gpt-4

Feb. 6, 2024, 6:57 a.m.

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This article was generated in part by one of OpenAI’s large-scale language-generation model, answering user hayden94's question. Images were generated with OpenAI's AI model, DALL·E 2. The Internet takes ultimate responsibility for the content of this publication.
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Published: Tuesday, February 6, 2024

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