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aküarios

  • one year ago

how to use "In"?

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  1. aküarios
    • one year ago
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    I have this question.. and I know i Have to use In... but dont know how Brooklyn has a goal to save $8,000 to buy a new entertainment system. In order to meet that goal, she deposited $4,132.79 into a savings account. If the account has an interest rate of 4.8% compounded quarterly, approximately when will Brooklyn be able to make the purchase?

  2. aküarios
    • one year ago
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    8000 = 4132.79(1+.048/12)^12*t 8000 = 4132.79(1.004)^12t (now I get stuck here..

  3. satellite73
    • one year ago
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    \[4132.79\times (1+\frac{.048}{4})^{4t}=8,000\]

  4. satellite73
    • one year ago
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    says QUARTERLY so be careful usually it is monthly

  5. aküarios
    • one year ago
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    yah u right... misread

  6. aküarios
    • one year ago
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    but then how do you continue... because that's when I get lost :/

  7. satellite73
    • one year ago
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    divide by 4132.78

  8. satellite73
    • one year ago
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    then use the change of base formula \[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]

  9. aküarios
    • one year ago
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    what actually In use for... like, I understand nothing at all about In... (In=log right) ?

  10. satellite73
    • one year ago
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    \[4132.79\times (1+\frac{.048}{4})^{4t}=8,000\] \[4132.79\times (1.012)^{4t}=8,000\] \[1.012^{4t}=8,000\div 4132.79\]

  11. aküarios
    • one year ago
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    and why is it use in this kind of formulas?

  12. satellite73
    • one year ago
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    \[1.012^{4t}=1.936\] rounded

  13. satellite73
    • one year ago
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    the to solve for \(4t\) use \[4t=\frac{\log(1.936)}{\log(1.012)}\]

  14. satellite73
    • one year ago
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    it doesn't matter what log you use

  15. aküarios
    • one year ago
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    mmm... but why log? like.. what is its function? I never understood when the teacher explained it...

  16. satellite73
    • one year ago
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    then i am sure i cannot explain it in a chat box here but basically \[b^x=y\iff \log_b(y)=x\]

  17. aküarios
    • one year ago
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    thats the formula right?

  18. satellite73
    • one year ago
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    it is a way to solve for a variable that is in the exponent

  19. aküarios
    • one year ago
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    mmmm.. got it.. the answer I got for the formula is t= 10.8 (then years and 8 month?)

  20. satellite73
    • one year ago
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    but you only have two logs on your calculator, \(\log_{10}(x)\) log base ten and \[\ln(x)=\log_e(x)\] which is log base e so if you want an actual decimal for an answer, you have to use \[b^x=A\iff x=\frac{\ln(A)}{\ln(b)}\]

  21. aküarios
    • one year ago
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    in my calculator i have log2, log10, and In... can use any true? (I used In)

  22. satellite73
    • one year ago
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    in other words, to solve for the variable in the exponent, it is the log of the total divided by the log of the base that is how to solve for a variable that is in the sky

  23. aküarios
    • one year ago
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    haha got it... so my answer is 10.8... how do i convert that into years and months?

  24. aküarios
    • one year ago
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    sorry ... i got 13.8

  25. satellite73
    • one year ago
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    makes no difference, all logs are the same \[\log_b(x)=\frac{\log_a(x)}{\log_a(b)}\]

  26. aküarios
    • one year ago
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    got it... but how do you convert t = 13.85 into years and months?

  27. satellite73
    • one year ago
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    what does \(t\) represent?

  28. aküarios
    • one year ago
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    time...

  29. aküarios
    • one year ago
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    so im guessing it would be 13 years and 8 months right?

  30. satellite73
    • one year ago
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    in what units?

  31. aküarios
    • one year ago
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    years&months

  32. satellite73
    • one year ago
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    yes t is time in years

  33. aküarios
    • one year ago
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    but in the choices I've got there is no 13 years and 8 month... (they have 13 years and 10 months) so I'm guessing is that one.. but they also have 13 years and 5 months... so like, i want to know how to solve it correctly...

  34. murrcat
    • one year ago
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    Was 13 years and 10 months the correct answer?

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