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BarlowGirl Group Title

PLEASE HELP!!!!!! URGENT!!!! When there's a table of information, for two different variables (like f(x) and x), how do I find the constant ratio to determine if it's an exponential function? Thank you SO much! - I really appreciate the help! =)

  • 2 years ago
  • 2 years ago

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  1. jim_thompson5910 Group Title
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    Do you have a table to use as an example?

    • 2 years ago
  2. jim_thompson5910 Group Title
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    If not, I can make one up if you want.

    • 2 years ago
  3. BarlowGirl Group Title
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    Sure, let me give an example . . . .

    • 2 years ago
  4. jim_thompson5910 Group Title
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    alright

    • 2 years ago
  5. BarlowGirl Group Title
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    |dw:1338949389152:dw|

    • 2 years ago
  6. BarlowGirl Group Title
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    Note: I made this up, so it most definitely won't have a constant ratio.

    • 2 years ago
  7. jim_thompson5910 Group Title
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    If this was an exponential equation, then it would be of the form y = a*b^x where a and b are some constant numbers

    • 2 years ago
  8. jim_thompson5910 Group Title
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    When x = 1, y is y = 1. So y = a*b^x 1 = a*b^1 1 = a*b Now solve for b to get b = 1/a

    • 2 years ago
  9. jim_thompson5910 Group Title
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    When x = 6, y is y = 5 So y = a*b^x 5 = a*b^6 Now plug in b = 1/a to get 5 = a*(1/a)^6 5 = a/(a^6) 5 = 1/a^5 5a^5 = 1 a^5 = 1/5 a = fifth root of 1/5 a = 0.72477 Since b = 1/a, we know that b = 1/a b = 1/0.72477 b = 1.3797 Therefore, if this was an exponential equation (it's probably not, but let's say it is), then the equation would be approximately y = 0.72477*1.3797^x

    • 2 years ago
  10. BarlowGirl Group Title
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    Okay, I'm following.

    • 2 years ago
  11. jim_thompson5910 Group Title
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    Now let's test another ordered pair. Let's test (8,4). Plug in x = 8 and y = 4 to get y = 0.72477*1.3797^x 4 = 0.72477*1.3797^4 4 = 0.72477*3.62358670182697 4 = 2.62626693388313 and we're way off So because of this, the table above is NOT modelling an exponential function

    • 2 years ago
  12. BarlowGirl Group Title
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    Okay, I think I understand. Thank you very much - you've been so helpful! :)

    • 2 years ago
  13. jim_thompson5910 Group Title
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    you're welcome, I'm glad I could help

    • 2 years ago
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