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Megaforte8600 Group Title

ln 2 - ln (3x + 2) = 1 Find X

  • 8 months ago
  • 8 months ago

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  1. mathmale Group Title
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    An important rule of logarithms is this : ln a - ln b = ln (a/b). Given that, what is ln 2 - ln (3x+2)?

    • 8 months ago
  2. Megaforte8600 Group Title
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    ln 2/ (3x + 2)

    • 8 months ago
  3. mathmale Group Title
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    good. Now, \[\ln \frac{ 2 }{ 3x+2 }=1.\] how would you now get rid of the ln operator? We want to solve this equation for x.

    • 8 months ago
  4. Megaforte8600 Group Title
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    e^ln ( 2/ 3x + 2 ) = e^1

    • 8 months ago
  5. mathmale Group Title
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    Hey, that's really cool. Can you now simplify this equation?

    • 8 months ago
  6. Megaforte8600 Group Title
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    no

    • 8 months ago
  7. Megaforte8600 Group Title
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    i forgot how to

    • 8 months ago
  8. mathmale Group Title
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    Use the fact that e^(ln a) = a. If this is true (which it is), then what is \[e ^{\ln \frac{ 2 }{ 3x+2 }}?\]

    • 8 months ago
  9. mathmale Group Title
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    The right side is simply e^1, or e.

    • 8 months ago
  10. Megaforte8600 Group Title
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    ok, so its now (2/3x + 2) = 1

    • 8 months ago
  11. mathmale Group Title
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    Other than that your 3x+1 should be within parentheses, yes! Mind re-writing that? Don't need parenthesis in front of the first 2.

    • 8 months ago
  12. Megaforte8600 Group Title
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    2/ (3x + 2) = e

    • 8 months ago
  13. mathmale Group Title
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    Yes, or \[\frac{ 3x+2 }{ 2 }=\frac{ 1 }{ e }\] One way of solving this would be to cross-multiply. We have to isolate x on the left side of the equation. I need to log off now, but think you're well on the way to solving this problem properly. See you again soon, I hope.

    • 8 months ago
  14. Megaforte8600 Group Title
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    im still confused

    • 8 months ago
  15. myininaya Group Title
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    are you confused on the cross multiplication part? he means if you have \[\frac{a}{b}=\frac{c}{d} \text{ you can do } ad=cb\]

    • 8 months ago
  16. myininaya Group Title
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    and remember we are solving for x remember treat e like you would treat 2 or 3 or any other constant number

    • 8 months ago
  17. Megaforte8600 Group Title
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    ok ^_^ thanks

    • 8 months ago
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