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anonymous

  • 5 years ago

PreCal please help! A population of bacteria is growing according to the equation . Estimate when the population will exceed 493. P(t)=400e^(0.05t)

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  1. anonymous
    • 5 years ago
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    So I know my equation is 493=400e^(0.05t) but I can't figure out how to solve it

  2. anonymous
    • 5 years ago
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    493 = 400*e^(0.05t) \[\ln (493/400) = 0.05t\]\[t=20*\ln (493/400) = 4.18\]

  3. anonymous
    • 5 years ago
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    where did you get 20 from? we haven't learned about natural log yet. I think that's what ln is...

  4. anonymous
    • 5 years ago
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    You are correct, ln is natural log :D It's just a special form of a logarithm with base "e" 1/.05 is 20 :D

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