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swiftskier96 Group Title

How do you solve this? Solve for k. 2000 = 3500e^(k*2)

  • one year ago
  • one year ago

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  1. swiftskier96 Group Title
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    Solve for k. 2000 = 3500e^(k*2)

    • one year ago
  2. swiftskier96 Group Title
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    @Frostbite @phi

    • one year ago
  3. Frostbite Group Title
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    Lets see: 2000 = 3500'e^(k*2) ln(2000) = ln(3500*e^(k*2)) ln(2000) = ln(3500) + ln(e^k*2) = ln(3500) + 2*k (ln(2000)-ln(3500))/2=k

    • one year ago
  4. Frostbite Group Title
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    then to make it perhaps a bit more easy: (ln(2000)-ln(3500))/2=k (ln(2000)-ln(3500))/2=k <-> ln(2000/3500)/2=k k=ln(4/7)/2

    • one year ago
  5. swiftskier96 Group Title
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    So k = In(4/7)/2 is the final answer?

    • one year ago
  6. phi Group Title
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    you can test your answer using a calculator

    • one year ago
  7. Frostbite Group Title
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    I just toke it on the gefüle..

    • one year ago
  8. phi Group Title
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    personally I would start with \[ 2000 = 3500e^{2k} \] and divide both sides by 3500 and simplify \[ \frac{4}{7} = e^{2k} \] then take the natural log of both sides \[ \ln\left(\frac{4}{7}\right) = 2k\] divide both sides by 2 to get k

    • one year ago
  9. swiftskier96 Group Title
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    Ok cool! Thanks both of you!! :)

    • one year ago
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