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animalsavior94

  • 3 years ago

condense this to a single quantity: 1/2lnX +ln(x+3)-ln(x^2 +1) thanks.

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  1. Owlfred
    • 3 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. LagrangeSon678
    • 3 years ago
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    This appears to become a quadratic equation

  3. animalsavior94
    • 3 years ago
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    really? well im not sure because its logarithms and u need to condense it.

  4. animalsavior94
    • 3 years ago
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    how did you get that?

  5. satellite73
    • 3 years ago
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    well it is wrong, so don't write it!

  6. satellite73
    • 3 years ago
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    i made a mistake

  7. animalsavior94
    • 3 years ago
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    oh i see could you please explain it to me?

  8. satellite73
    • 3 years ago
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    \[\frac{1}{2}\ln(x)=\ln(\sqrt{x})\]

  9. satellite73
    • 3 years ago
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    because \[\ln(x^n)=n\ln(x)\]

  10. satellite73
    • 3 years ago
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    here i am going from the right hand side to the left hand side.

  11. satellite73
    • 3 years ago
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    \[\frac{1}{2}\ln(x)=\ln(x^{\frac{1}{2}})=\ln(\sqrt{x})\]

  12. satellite73
    • 3 years ago
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    then we use \[\ln(ab) = \ln(a)+ln(b)\] again from the right to the left.

  13. satellite73
    • 3 years ago
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    so we get \[\ln(\sqrt{x})+\ln(x+3)=\ln(\sqrt{x}(x+3))\]

  14. satellite73
    • 3 years ago
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    then we use \[\ln(a)-\ln(b)=\ln(\frac{a}{b})\]

  15. satellite73
    • 3 years ago
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    to get \[\ln(\sqrt{x}(x+3))-\ln(x^2+1)=\ln(\frac{\sqrt{x}(x+3)}{x^2+1})\]

  16. satellite73
    • 3 years ago
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    i made a mistake on the first try and wrote \[\frac{1}{2}\ln(x)=\ln(x^2)\] which is wrong

  17. satellite73
    • 3 years ago
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    there now i deleted it. hope the steps are clear

  18. animalsavior94
    • 3 years ago
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    ohhh no i understand thanks so much!!:]

  19. satellite73
    • 3 years ago
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    welcome!

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