\( 2 = e^{200k} \Rightarrow k = \frac{\ln 2}{200} \)

February 22, 2026

$2 = e^{200k} \Rightarrow k = \frac{\ln 2}{200}$

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A palynologist analyzing sediment layers finds that the pollen count of oak trees increases exponentially over time, doubling every 200 years. If the current count in layer A (1000 years old) is 48 grains per gram, what was the count 600 years ago, assuming continuous exponential growth?
The formula for exponential growth is: \( P(t) = P_0 \cdot e^{kt} \)
We know that \( P(200) = 2P_0 \), so:
Now, we go 600 years before 1000 years ago, so 800 years before layer A:
\( P(-800) = P_0 \cdot e^{(\ln 2 / 200)(-800)} = P_0 \cdot e^{-4\ln 2} = P_0 \cdot (e^{\ln 2})^{-4} = P_0 \cdot 2^{-4} = P_0 / 16 \)
Given that current count (1000 years ago, t = 0) is 48, so \( P_0 = 48 \) (at t = 0 for layer A)

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