So \( r = \frac{0.4699}{6} \approx 0.0783 \) → 7.83%

February 22, 2026

$So \( r = \frac{0.4699}{6} \approx 0.0783 \) → 7.83%$

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In a carbon sequestration study, a forest absorbs CO₂ at a rate proportional to its current biomass. If biomass grows from 500 tons to 800 tons in 6 years, what is the approximate annual growth rate (as a percent), using the formula \( B(t) = B_0 e^{rt} \)?
\( 800 = 500 e^{6r} \Rightarrow \frac{8}{5} = 1.6 = e^{6r} \Rightarrow \ln(1.6) = 6r \)
\( \ln(1.6) = \ln(8/5) = \ln 8 - \ln 5 = 3\ln 2 - \ln(5) \approx 3(0.6931) - 1.6094 = 2.0793 - 1.6094 = 0.4699 \)
Rounded to nearest percent: 8%
But check: 500 × e^(0.0783×6) ≈ 500 × e^0.4698 ≈ 500 × 1.600 = 800 → correct
So 7.83% → closest whole percent is 8%

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