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greenandblue

  • 2 years ago

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.

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  1. bakonloverk
    • 2 years ago
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    multiply 1,500 * 0.03 * 5

  2. bakonloverk
    • 2 years ago
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    then add tht 2 1,500

  3. bakonloverk
    • 2 years ago
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    so it would b 1,500 + 225

  4. bakonloverk
    • 2 years ago
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    so the first answer is $1,725

  5. greenandblue
    • 2 years ago
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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?

  6. bakonloverk
    • 2 years ago
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    the 225 is 1,500 * 0.03 * 5

  7. bakonloverk
    • 2 years ago
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    and no i dont know wht the natural logarithm is

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

  9. bakonloverk
    • 2 years ago
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    lol ur welcome

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

  11. tkhunny
    • 2 years ago
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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 \(t\) \(t = \dfrac{ln(2)}{0.03} = 23.105\)

  12. tkhunny
    • 2 years ago
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    @greenandblue @bakonloverk @satellite73

  13. bakonloverk
    • 2 years ago
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    u called 4 me?

  14. tkhunny
    • 2 years ago
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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. :-)

  15. bakonloverk
    • 2 years ago
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    ohhhh kk

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