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AmTran_Bus

  • one year ago

Please help me solve for k!

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  1. AmTran_Bus
    • one year ago
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    |dw:1433563190186:dw|

  2. AmTran_Bus
    • one year ago
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    The book says k= 1/10 ln 3040/2560 But where did that come from?

  3. AmTran_Bus
    • one year ago
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    Is it just a plain ration between t and, well, k?

  4. zepdrix
    • one year ago
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    \[\Large\rm y(10)=2560e^{10k}=3040\]Apply log, then log rule :)

  5. zepdrix
    • one year ago
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    Err divide first, woops,\[\Large\rm \frac{3040}{2560}=e^{10k}\]

  6. zepdrix
    • one year ago
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    Log each side,\[\Large\rm \ln\left(\frac{3040}{2560}\right)=\ln\left(e^{10k}\right)\]

  7. zepdrix
    • one year ago
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    log rule,\[\Large\rm \ln\left(\frac{3040}{2560}\right)=10k\ln\left(e\right)\]Then uh, simplify...\[\Large\rm \ln\left(\frac{3040}{2560}\right)=10k\cdot 1\]ya?

  8. AmTran_Bus
    • one year ago
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    Yes. That is making some sense. So then, you have .178=10k

  9. zepdrix
    • one year ago
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    .1719* woops i think u missed a decimal value there.

  10. AmTran_Bus
    • one year ago
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    My bad! So, what do you do with the 10k?

  11. zepdrix
    • one year ago
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    solve for k 0_o 10 is multiplying the k, ya? do the inverse of multiplication to isolate your k

  12. AmTran_Bus
    • one year ago
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    I mean, it is set equal to it as is, but dont you plug it into k?

  13. AmTran_Bus
    • one year ago
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    Oh, your post above, sorry!

  14. AmTran_Bus
    • one year ago
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    .01719

  15. AmTran_Bus
    • one year ago
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    James Stewart wrote some of these examples confusing!

  16. zepdrix
    • one year ago
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    hehe

  17. AmTran_Bus
    • one year ago
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    Thank you.

  18. zepdrix
    • one year ago
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    no probsss

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