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RaphaelFilgueiras

  • 2 years ago

9^x-7.3^x+10=0,what is the minimum value wich made this equation correct?given(log2=0.3,log3=0.48)

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  1. RaphaelFilgueiras
    • 2 years ago
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    3^x=u get it!

  2. RaphaelFilgueiras
    • 2 years ago
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    u²-7.u+10=0

  3. experimentX
    • 2 years ago
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    Hmm ... was that 3^x ??not 7.3^x ??

  4. RaphaelFilgueiras
    • 2 years ago
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    yes my bad

  5. experimentX
    • 2 years ago
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    i was wondering it simply was untrue ... lol

  6. RaphaelFilgueiras
    • 2 years ago
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    3^x=2

  7. experimentX
    • 2 years ago
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    x = log 2 / log 3

  8. RaphaelFilgueiras
    • 2 years ago
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    x=.3/.48

  9. experimentX
    • 2 years ago
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    I bet the minimum value that would make this equation true would be http://www.wolframalpha.com/input/?i=x^2+-+6.3x+%2B+10+%3D+0+

  10. RaphaelFilgueiras
    • 2 years ago
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    the minimum is 0.625

  11. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=x^2+-+2+sqrt%2810%29+x+%2B+10+%3D+0+

  12. experimentX
    • 2 years ago
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    Hmmm ... is it asking for two values of x??

  13. RaphaelFilgueiras
    • 2 years ago
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    no only the minimum

  14. experimentX
    • 2 years ago
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    Oh ... i guess it is solved as like you did then

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