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anonymous

  • 5 years ago

Exponential equation. Points (-2,9) (0,1). My answer y=3(1/3)^x. Am I correct?

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  1. anonymous
    • 5 years ago
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    Well \[3*(1/3)^{-2} = 3*3^2 = 9*3 = 27 \ne 9\] so nope. That can't be right.

  2. anonymous
    • 5 years ago
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    hmm well lets see. First I got a=9/b^2. is that correct?

  3. anonymous
    • 5 years ago
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    From the second point we have \[1 = ab^0\] Since \[b^0=1\rightarrow 1 = ab^0 = a*1 \rightarrow a = 1\]

  4. anonymous
    • 5 years ago
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    Try plotting y = 2^x, y = 2^(-x), etc. That may give you ideas about the base and the exponent,

  5. anonymous
    • 5 years ago
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    oh well I was taught to use the first point first to find out what a equals for the b equation.

  6. anonymous
    • 5 years ago
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    ....? plotting?

  7. anonymous
    • 5 years ago
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    Anytime you have x = 0 for an exponential equation, the y value for that point is the value of a.

  8. anonymous
    • 5 years ago
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    Because any base raised to the 0 power equals 1.

  9. anonymous
    • 5 years ago
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    still no comprendo senor.....or senora. You don't use the second equation for a.......gawh I wish math was simple for fooligans like me.

  10. anonymous
    • 5 years ago
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    Ok. Lets break it down. We have 2 points (-2,9) and (0,1) That gives 2 equations: \[9 = ab^{-2}\] and \[1 = ab^0\]

  11. anonymous
    • 5 years ago
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    With me so far?

  12. anonymous
    • 5 years ago
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    si.

  13. anonymous
    • 5 years ago
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    Ok. So what is \[b^0=?\]

  14. anonymous
    • 5 years ago
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    this is where your throwing me off. Don't i have to use the first equation for a first?

  15. anonymous
    • 5 years ago
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    It doesn't matter. You use whichever is easier. In this case the second equation is easier. This will become apparent when you understand what b^0 is.

  16. anonymous
    • 5 years ago
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    okay. I just want to make sure. I have a test tomorrow and if i use your tricks I don't want to get penalized.

  17. anonymous
    • 5 years ago
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    Nope. Not a trick.

  18. anonymous
    • 5 years ago
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    okay well to me its a different way so continue senor.

  19. anonymous
    • 5 years ago
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    If you really want we can do it your way.

  20. anonymous
    • 5 years ago
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    no I'm open to your idea. your the smarter one here so continue.

  21. anonymous
    • 5 years ago
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    Lets do it your way and then we'll do mine and see that we get the same thing.

  22. anonymous
    • 5 years ago
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    okay sounds good to me. :0]

  23. anonymous
    • 5 years ago
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    Are you familiar with this notation? \[\rightarrow\]

  24. anonymous
    • 5 years ago
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    does that mean equals or references to the next step?

  25. anonymous
    • 5 years ago
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    not equals. It means 'implies', 'therefore', or 'leads to'

  26. anonymous
    • 5 years ago
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    So something like \[ A = 5\] \[ A +B = 10 \rightarrow 5 + B = 10 \rightarrow B = 5\]

  27. anonymous
    • 5 years ago
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    Make sense?

  28. anonymous
    • 5 years ago
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    okay continue :0]

  29. anonymous
    • 5 years ago
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    So using the first equation and solving for a: \[9 = ab^{-2}\] \[\rightarrow 9b^2 = ab^{-2}b^2\] \[\rightarrow9b^2 = ab^0 \rightarrow a = 9b^2\]

  30. anonymous
    • 5 years ago
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    Follow that?

  31. anonymous
    • 5 years ago
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    Okay question time! Don't you divide by b^2? and keep 9/b^2 for the next equations a? for example: 1=(9/b^2)(b^0)

  32. anonymous
    • 5 years ago
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    then solve for b and then plug in b for the original equation.

  33. anonymous
    • 5 years ago
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    Ah. No because \[b^{-2} = \frac{1}{b^2}\]

  34. anonymous
    • 5 years ago
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    So you need to multiply by b^2 in order to get a by itself, not divide.

  35. anonymous
    • 5 years ago
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    oooooh so thats where I missed the mark. Okay i get it so if i have a negative, instead of dividing I multiply.

  36. anonymous
    • 5 years ago
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    Think about it this way. You want to move the b over to the other side. You know that multiplying powers of the same base you add exponents right? \[a^b * a^c = a^{(b+c)}\]

  37. anonymous
    • 5 years ago
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    yes and when you divide you subtract.

  38. anonymous
    • 5 years ago
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    Right. So here's the million dollar question. What is \[k^0\]

  39. anonymous
    • 5 years ago
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    0 right?

  40. anonymous
    • 5 years ago
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    Nope. 2^4 = 16 2^3 = 8 2^2 = 4 2^1 = ? 2^0 = ?

  41. anonymous
    • 5 years ago
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    shoot I change my mind its 1!! now wheres my money?

  42. anonymous
    • 5 years ago
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    heheh.

  43. anonymous
    • 5 years ago
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    Right. So if we have \[ab^2 = 9\] We can multiply both sides by b^-2 \[a*b^2*b^{-2} = 9*b^{-2}\]

  44. anonymous
    • 5 years ago
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    And we end up with \[a*b^{2+ (-2)} = 9b^{-2} \] Which simplifies to \[a*b^{0} = 9b^{-2} \] Which simplifies to \[a*1 = 9b^{-2} \] Which simplifies to \[a = 9b^{-2} \]

  45. anonymous
    • 5 years ago
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    Now that wasn't the equation we had. We had \[ab^{-2} = 9\] So instead of multiplying by b^-2 we need to multiply by b^2

  46. anonymous
    • 5 years ago
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    okay gotcha.

  47. anonymous
    • 5 years ago
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    So we end up with \[a = 9b^2\]

  48. anonymous
    • 5 years ago
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    And we plug that into the second equation.

  49. anonymous
    • 5 years ago
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    So what do you get when you plug in 9b^2 for a in the second equation.

  50. anonymous
    • 5 years ago
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    i got 1/9=b^2

  51. anonymous
    • 5 years ago
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    after dividing 9 from 9b^2 of course.

  52. anonymous
    • 5 years ago
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    So b is?

  53. anonymous
    • 5 years ago
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    b=1/3 Like I had before...

  54. anonymous
    • 5 years ago
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    Yeah. So now plug that in for b in either of the two equations.

  55. anonymous
    • 5 years ago
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    then i got a=1. and a full answer of y=1(1/3)^x

  56. anonymous
    • 5 years ago
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    Yep.

  57. anonymous
    • 5 years ago
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    So now look at what happens when you plug in 1/3 for b in the second equation. Does it matter what b was?

  58. anonymous
    • 5 years ago
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    what do you mean? I got the whole equation right didnt i?

  59. anonymous
    • 5 years ago
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    Yes. I'm trying to show my original idea.

  60. anonymous
    • 5 years ago
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    oh okay. well no it doesn't matter what b was. So yes you were right. hahaa that all you wanted to hear wasnt it?

  61. anonymous
    • 5 years ago
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    \[1=ab^0\] is our second equation from the beginning. You plug in 1/3 and you get a = 1. But does it matter what you plug in for b?

  62. anonymous
    • 5 years ago
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    no just as long as the answer is correct. :0]

  63. anonymous
    • 5 years ago
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    Right. So you can see that for exponential equations if you are given a point (0,k). You already know that a must be k.

  64. anonymous
    • 5 years ago
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    oh okay yeah I get it.

  65. anonymous
    • 5 years ago
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    Because \[ab^0= k\rightarrow a*1 = k \rightarrow a = k\]

  66. anonymous
    • 5 years ago
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    Which means you can jump right to finding b and save some time.

  67. anonymous
    • 5 years ago
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    so if i do it either way i will end up with the right answer. Okay i'm getting your drift here. thanks a bunch! your so patient and I feel more confident it passing my test. SO thanks it really means a lot!!

  68. anonymous
    • 5 years ago
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    Anytime =)

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