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YesterdayiSaidTomorow

  • 2 years ago

Use the properties of logarithms to expand the expression ln z(z-1)^2, z>1?

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  1. .UserNotFound.
    • 2 years ago
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    Check it on yahoo . It says " = Ln (z) + Ln (z-1)^2 = Ln (z) + 2 Ln (z-1) "

  2. YesterdayiSaidTomorow
    • 2 years ago
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    that's the question I posted

  3. YesterdayiSaidTomorow
    • 2 years ago
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    and no one would explain

  4. .UserNotFound.
    • 2 years ago
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    amofide xD

  5. amorfide
    • 2 years ago
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    you have rules for logs if you have lnx² that is the same as saying 2lnx you can bring the power to the front of the log

  6. YesterdayiSaidTomorow
    • 2 years ago
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    but how do you separate it? Like Ln (z) + Ln (z-1)^2

  7. YesterdayiSaidTomorow
    • 2 years ago
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    i mean Ln (z) + 2 Ln (z-1)

  8. amorfide
    • 2 years ago
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    also you started with ln z(z-1)^2 there is another rule for logs if you have lna + ln b this is equal to lnab so you have ln z(z-1)^2 this is being multiplied, so we know we have two logs added together lnz + ln (z-1)²

  9. YesterdayiSaidTomorow
    • 2 years ago
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    Ooooooooooooooh

  10. YesterdayiSaidTomorow
    • 2 years ago
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    Thanks!

  11. amorfide
    • 2 years ago
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    anytime :D so lna+lnb=lnab lna-lnb=lna/b lna^b=blna i think there is another one i forget...

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