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AonZ

  • 2 years ago

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

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  1. AonZ
    • 2 years ago
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    check the identity by evaluating...

  2. AonZ
    • 2 years ago
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    \[\huge \log_{\frac{ 1 }{ 5 }}25 \]

  3. hartnn
    • 2 years ago
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    use the same property i gave for last Q

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

  5. AonZ
    • 2 years ago
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    |dw:1360930788428:dw|

  6. hartnn
    • 2 years ago
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    yes, simplify numerator and denominator separately.

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

  8. ParthKohli
    • 2 years ago
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    Why so serious?

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

  10. AonZ
    • 2 years ago
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    ahh @agent0smith explained it pretty good :)

  11. AonZ
    • 2 years ago
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    @agent0smith How did you get the base to be 5? Like all of them?

  12. agent0smith
    • 2 years ago
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    Not sure what you mean... I just used base 5 because it makes it easy to solve.

  13. ParthKohli
    • 2 years ago
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    @AonZ Change of bases.

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

  15. AonZ
    • 2 years ago
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    so if you use change of base rule you can choose ANY base?

  16. ParthKohli
    • 2 years ago
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    Yes.

  17. AonZ
    • 2 years ago
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    wow i reckon i understand logs a lot better now :P

  18. agent0smith
    • 2 years ago
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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"?

  19. ParthKohli
    • 2 years ago
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    Yes, I am sorry.

  20. ParthKohli
    • 2 years ago
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    But you get the point.

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

  22. ParthKohli
    • 2 years ago
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    :-)

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

  24. hartnn
    • 2 years ago
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    or simply \(\huge \log_{\frac{ 1 }{ 5 }} 5 = \frac{ \log_{} 5 }{ \log_{}\frac{ 1 }{ 5}} \)

  25. ParthKohli
    • 2 years ago
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    Much better up there.

  26. hartnn
    • 2 years ago
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    \(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}\)

  27. agent0smith
    • 2 years ago
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    ^ that works fine, but requires a calculator.

  28. ParthKohli
    • 2 years ago
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    Nope.

  29. hartnn
    • 2 years ago
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    \(\huge \log_{\frac{ 1 }{ 5 }} 25 = \frac{ \log_{} 25 }{ \log_{}\frac{ 1 }{ 5}}= \frac{ \log_{} 5^2 }{ \log_{}\frac{ 1 }{ 5}}\)

  30. hartnn
    • 2 years ago
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    calcy for what ?

  31. ParthKohli
    • 2 years ago
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    \[\dfrac{\log _5 5^2}{\log _5 5^{-1}} = \]

  32. agent0smith
    • 2 years ago
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    I'm assuming @hartnn is using base 10...

  33. ParthKohli
    • 2 years ago
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    \[\dfrac{2}{-1}\]

  34. hartnn
    • 2 years ago
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    does not matter what base i use, by default its 10

  35. ParthKohli
    • 2 years ago
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    You can use change of base too.

  36. agent0smith
    • 2 years ago
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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?

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

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

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