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Home » Multiply the initial population by \( 2^8 \): - iBuildNew

Multiply the initial population by \( 2^8 \): - iBuildNew

📅 April 29, 2026 👤 scraface

Apr 29, 2026

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\frac{1000}{900} > 1 + 9e^{-0.2t} \Rightarrow \frac{10}{9} > 1 + 9e^{-...
\frac{10}{9} - 1 > 9e^{-0.2t} \Rightarrow \frac{1}{9} > 9e^{-0.2t}
\frac{1}{81} > e^{-0.2t}
Take natural log:
\ln\left(\frac{1}{81}\right) > -0.2t \Rightarrow -\ln(81) > -0.2t
\ln(81) < 0.2t
\( \ln(81) = \ln(3^4) = 4\ln(3) \approx 4 \times 1.0986 = 4.3944 \)
< 0.2t \Rightarrow t > \frac{4.3944}{0.2} = 21.972
So the smallest integer \( t \
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