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Adorkable

  • 3 years ago

T(t)=63(.05)^(t/10)+19 . Where T is temperature, and t is time in minutes. How long did it take for a sandwich to reach an internal temputure of 30 degrees. I know that I have to substitute 30 into T. But I don't know how to solve for t. Please help and show steps.

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  1. UF_Cormac
    • 3 years ago
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    30 = 63(0.5)^(t/10) + 19 11 = 63(0.5)^(t/10) 11/63 = (0.5)6(t/10) ln(11/63) = (t/10)ln(0.5) Should be able to solve from there.

  2. Adorkable
    • 3 years ago
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    I'm not able to solve from here. How would I do this. Please.

  3. Adorkable
    • 3 years ago
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    I get where you got everything from what I"m having trouble with is isolating t.

  4. Adorkable
    • 3 years ago
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    As of right now I have 11/63=(.5)^t/10

  5. UF_Cormac
    • 3 years ago
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    Do you know logarithms?

  6. UF_Cormac
    • 3 years ago
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    Or at least what they are?

  7. Adorkable
    • 3 years ago
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    Nope. I think we learn that next year.

  8. jdoe0001
    • 3 years ago
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    makes me wonder why you're doing this now then :|

  9. UF_Cormac
    • 3 years ago
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    Agreed, the only way to solve for the independent variable t is to use logarithms.

  10. Adorkable
    • 3 years ago
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    Lol, Can you give me the jist of this? I probably have learned it but we don't call it logarithms.

  11. UF_Cormac
    • 3 years ago
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    http://www.purplemath.com/modules/logs.htm

  12. jdoe0001
    • 3 years ago
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    so, you learned it "next year", but can't recall it in the present..... ooookkk

  13. Adorkable
    • 3 years ago
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    I haven't learned it. Its in next years curriculum im told...

  14. Adorkable
    • 3 years ago
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    We havent learned this. o.o Log business.

  15. jdoe0001
    • 3 years ago
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    so, this exercise doesn't apply then

  16. UF_Cormac
    • 3 years ago
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    \[\log_{b}y = x ...... b^x=y \]

  17. Adorkable
    • 3 years ago
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    Thanks for your help I guess.

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