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AonZ

  • 3 years ago

Log question. Is there a rule that can solve this? \[\huge \log_{\frac{ 1 }{ a }} \frac{ 1 }{ n }\]

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  1. AonZ
    • 3 years ago
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    @agent0smith can you help with this?

  2. agent0smith
    • 3 years ago
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    Yeah i'm not sure what you want to do with it... you can change it into a couple of different forms?

  3. yakzo
    • 3 years ago
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    There is a formula for changing the base of the logarithm.

  4. AonZ
    • 3 years ago
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    well il type full equation prove log a(x) = -log 1/2 (x)

  5. hartnn
    • 3 years ago
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    use : \(\huge \log_ab=\dfrac{\log b}{\log a}\)

  6. hartnn
    • 3 years ago
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    u sure, this is the Q ? prove log a(x) = -log 1/2 (x)

  7. AonZ
    • 3 years ago
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    wait my bad LOL log a(x) = -log 1/a (x)

  8. hartnn
    • 3 years ago
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    right, use the property i gave on RIGHT side.

  9. AonZ
    • 3 years ago
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    |dw:1360928528477:dw|

  10. AonZ
    • 3 years ago
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    |dw:1360928591721:dw| these cancel right?

  11. hartnn
    • 3 years ago
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    simplify both numerator and denominator. and - log x = log(1/x) was unnecessary and no, not directly.

  12. AonZ
    • 3 years ago
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    how else do i do it then??

  13. hartnn
    • 3 years ago
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    |dw:1360928748794:dw|

  14. AonZ
    • 3 years ago
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    oh so the negatives cancel?

  15. hartnn
    • 3 years ago
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    yesh, now negatives can cancel.

  16. hartnn
    • 3 years ago
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    again apply that property on what you have...

  17. hartnn
    • 3 years ago
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    \(\huge \dfrac{\log b}{\log a}=\log_ab\)

  18. AonZ
    • 3 years ago
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    yes i can see how the equal each other now :)

  19. agent0smith
    • 3 years ago
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    \[\log_{a} x = -\log_{\frac{ 1 }{ a }} x\] Use the log rule on the right side \[-\log_{\frac{ 1 }{ a }} x =- \frac{ \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } =- \frac{ \log_{a} x }{ \log_{a} a ^{-1}}=- \frac{ \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}\]

  20. agent0smith
    • 3 years ago
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    moved the negatives, makes it easier to follow. \[-\log_{\frac{ 1 }{ a }} x = \frac{ - \log_{a} x }{ \log_{a} \frac{ 1 }{ a } } = \frac{- \log_{a} x }{ \log_{a} a ^{-1}}= \frac{- \log_{a} x }{- \log_{a} a} = \frac{ \log_{a} x }{1}\]

  21. AonZ
    • 3 years ago
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    ahh ok thanks to all of you :)

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