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greenandblue

PLEASE HELP ME ON THIS!! Joe deposits $1,500 in an account that pays 3% annual interest compounded continuously. a. How much will Joe have in his account after 5 years? b. How long will it take Joe to double his money? Use natural logarithms and explain your answer.

  • one year ago
  • one year ago

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  1. bakonloverk
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    multiply 1,500 * 0.03 * 5

    • one year ago
  2. bakonloverk
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    then add tht 2 1,500

    • one year ago
  3. bakonloverk
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    so it would b 1,500 + 225

    • one year ago
  4. bakonloverk
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    so the first answer is $1,725

    • one year ago
  5. greenandblue
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    Thank you so much but i got one question how did you get the 225 and also what does it mean natural logarithm?

    • one year ago
  6. bakonloverk
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    the 225 is 1,500 * 0.03 * 5

    • one year ago
  7. bakonloverk
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    and no i dont know wht the natural logarithm is

    • one year ago
  8. greenandblue
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    okay thank you so much seriously you have no idea!:)

    • one year ago
  9. bakonloverk
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    lol ur welcome

    • one year ago
  10. satellite73
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    \[1500(1.03)^5\] for the first one second one solve \[2=(1.03)^t\] via \[t=\frac{\ln(2)}{\ln(1.03)}\]

    • one year ago
  11. tkhunny
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    We seem to be missing an important word in the original problem statement. "CONTINUOUSLY." a) \(1500\cdot e^{0.03\cdot5} = 1742.75\) b) \(1500\cdot e^{0.03\cdot t} = 2\cdot 1500 = 3000\) or \(e^{0.03\cdot t} = 2\) Solving for \(t\) \(t = \dfrac{ln(2)}{0.03} = 23.105\)

    • one year ago
  12. tkhunny
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    @greenandblue @bakonloverk @satellite73

    • one year ago
  13. bakonloverk
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    u called 4 me?

    • one year ago
  14. tkhunny
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    Yup. You did Simple Interest, Satellite73 did Annual Compound Interest. The problem statement requires Continuous Compounding. That's all. Lot's of ways to tackle these things. That's why banks and insurance companies are so confusing. :-)

    • one year ago
  15. bakonloverk
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    ohhhh kk

    • one year ago
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