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Quadratic Exponential question

  1. Jun 12, 2005 #1
    This was on a test but I couldn't quite solve for x:

    [tex]5^2^x + 4(5)^x = -3[/tex]

    let [itex]5^x = y[/itex]

    [tex]y^2 + 4y + 3 = 0[/tex]
    [tex](y + 1)(y + 3) = 0[/tex]

    So I end up with [itex]5^x = -1[/itex] or [itex]5^x = -3[/itex], but I don't think that makes sense... what am I doing wrong? It must be something dumb I'm doing :shy:
     
    Last edited: Jun 12, 2005
  2. jcsd
  3. Jun 12, 2005 #2

    Hurkyl

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    Maybe you haven't done anything wrong? What would that imply?
     
  4. Jun 12, 2005 #3
    Ehh I was kind of worried about that... I guess it could imply I suck at logs? :tongue:

    I would continue by taking the log of both sides, but as far as I know you cannot take the log of a negative number... my calculator agrees with me. :biggrin:

    Overall, I don't know what this means.
     
    Last edited: Jun 12, 2005
  5. Jun 12, 2005 #4
    It meons there's no real number x that is the root of this function... Think of it this way, which power of 5(a positive number) would give you -1 or -3?
     
  6. Jun 12, 2005 #5
    How would I work it out to get an answer? All we've been taught on complex numbers (if that's what you're implying) is that i is the square root of negative one.
     
  7. Jun 12, 2005 #6
    What he's saying is that there is no value for x that satisfies that equation.
     
  8. Jun 12, 2005 #7
    Why would this be on my grade 11 math exam then? :mad:
     
  9. Jun 12, 2005 #8
    I don't know, but surely you can see that there's no power x that you can raise 5 to to get a negative number, not even a negative one. Your working at the top was correct.
     
  10. Jun 12, 2005 #9

    saltydog

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    Well it's an interesting problem for me. There are no solutions with "real numbers" but there are two complex-number solutions which perhaps it's best to not worry about now unless you want to know how to find them.
     
  11. Jun 12, 2005 #10

    Hurkyl

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    Presumably to test if you can identify when equations have no solutions. :tongue2:
     
  12. Jun 12, 2005 #11
    Crazy! Anyway, thanks for all your help.
     
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