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AonZ Group Title

Last question!!! log 1/5 (25)

  • one year ago
  • one year ago

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  1. AonZ Group Title
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    check the identity by evaluating...

    • one year ago
  2. AonZ Group Title
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    \[\huge \log_{\frac{ 1 }{ 5 }}25 \]

    • one year ago
  3. hartnn Group Title
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    use the same property i gave for last Q

    • one year ago
  4. ParthKohli Group Title
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    \(\frac{1}{5} = 5^{-1}\). What would you multiply to that to get \(5^2\)?

    • one year ago
  5. AonZ Group Title
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    |dw:1360930788428:dw|

    • one year ago
  6. hartnn Group Title
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    yes, simplify numerator and denominator separately.

    • one year ago
  7. agent0smith Group Title
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    \[\large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{5} 25 }{ \log_{5}\frac{ 1 }{ 5}} = \frac{ \log_{5} 5^2 }{ \log_{5} 5^{-1}}\]

    • one year ago
  8. ParthKohli Group Title
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    Why so serious?

    • one year ago
  9. ParthKohli Group Title
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    Just figure out what you multiply to \(5^{-1}\) to get \(5^2.\)

    • one year ago
  10. AonZ Group Title
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    ahh @agent0smith explained it pretty good :)

    • one year ago
  11. AonZ Group Title
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    @agent0smith How did you get the base to be 5? Like all of them?

    • one year ago
  12. agent0smith Group Title
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    Not sure what you mean... I just used base 5 because it makes it easy to solve.

    • one year ago
  13. ParthKohli Group Title
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    @AonZ Change of bases.

    • one year ago
  14. hartnn Group Title
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    you can choose any base for using that property,e,5,10,... thats why its unspecified in the property

    • one year ago
  15. AonZ Group Title
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    so if you use change of base rule you can choose ANY base?

    • one year ago
  16. ParthKohli Group Title
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    Yes.

    • one year ago
  17. AonZ Group Title
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    wow i reckon i understand logs a lot better now :P

    • one year ago
  18. agent0smith Group Title
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    ^ correct @ParthKohli shouldn't it be "figure out what power you need to raise 5^-1 to, to get 25" as opposed to "Just figure out what you multiply to 5^−1 to get 5^2"?

    • one year ago
  19. ParthKohli Group Title
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    Yes, I am sorry.

    • one year ago
  20. ParthKohli Group Title
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    But you get the point.

    • one year ago
  21. agent0smith Group Title
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    Ah okay, cos i was wondering how this helps:\[5^{-1} \times 5^3 = 25\]compared to this \[(5^{-1})^{-2}=25\]

    • one year ago
  22. ParthKohli Group Title
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    :-)

    • one year ago
  23. agent0smith Group Title
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    @AonZ you could also do it this way (but it makes it a bit more difficult): \[ \large \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{25} 25 }{ \log_{25}\frac{ 1 }{ 5}} = \frac{1 }{ \log_{25} 25^{-0.5}}\]

    • one year ago
  24. hartnn Group Title
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    or simply \(\huge \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{} 5 }{ \log_{}\frac{ 1 }{ 5}} \)

    • one year ago
  25. ParthKohli Group Title
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    Much better up there.

    • one year ago
  26. hartnn Group Title
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    \(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}\)

    • one year ago
  27. agent0smith Group Title
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    ^ that works fine, but requires a calculator.

    • one year ago
  28. ParthKohli Group Title
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    Nope.

    • one year ago
  29. hartnn Group Title
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    \(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}\)

    • one year ago
  30. hartnn Group Title
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    calcy for what ?

    • one year ago
  31. ParthKohli Group Title
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    \[\dfrac{\log _5 5^2}{\log _5 5^{-1}} = \]

    • one year ago
  32. agent0smith Group Title
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    I'm assuming @hartnn is using base 10...

    • one year ago
  33. ParthKohli Group Title
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    \[\dfrac{2}{-1}\]

    • one year ago
  34. hartnn Group Title
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    does not matter what base i use, by default its 10

    • one year ago
  35. ParthKohli Group Title
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    You can use change of base too.

    • one year ago
  36. agent0smith Group Title
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    \[ \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}} \] I mean, how are you solving this w/o changing base again, or using a calculator?

    • one year ago
  37. hartnn Group Title
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    \(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}=\dfrac{2 \log 5}{-1 \log 5}=-2\)

    • one year ago
  38. agent0smith Group Title
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    Oh you're simplifying to \[ \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}} = \frac{ 2 \log5 }{ -\log5 }\]

    • one year ago
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