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bronzegoddess Group Title

prove the rule:

  • 2 years ago
  • 2 years ago

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  1. bronzegoddess Group Title
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    \[a^{x}=\exp(xlna)\]

    • 2 years ago
  2. bronzegoddess Group Title
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    and \[\log _{a}=\frac{ lnx }{ lna }\]

    • 2 years ago
  3. henpen Group Title
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    \[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]

    • 2 years ago
  4. henpen Group Title
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    For the second, do you mean\[\log_a x=\frac{lnx}{lna}\]?

    • 2 years ago
  5. henpen Group Title
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    \[\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}\]

    • 2 years ago
  6. henpen Group Title
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    \[\large \ln x^{\log_a e} ln a= lnx\] \[\large ( x^{\log_a e} + a)=x\]

    • 2 years ago
  7. henpen Group Title
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    \[\large x^{\log_a e}=x-a\] \[\large x^{\log_a e}=x^{log_x{(x-a)}}\] \[\large {\log_a e}={log_x{x-a}}\]

    • 2 years ago
  8. bronzegoddess Group Title
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    @henpen , yes sorry i forgot the x

    • 2 years ago
  9. henpen Group Title
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    I'm not quite sure where to go after that

    • 2 years ago
  10. bronzegoddess Group Title
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    am not sure i follow ur steps.. how did you get all the those terms in the first part?

    • 2 years ago
  11. henpen Group Title
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    |dw:1352237942749:dw|

    • 2 years ago
  12. henpen Group Title
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    |dw:1352237963548:dw|

    • 2 years ago
  13. henpen Group Title
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    Or...|dw:1352238104786:dw|

    • 2 years ago
  14. bronzegoddess Group Title
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    my problem isnt that i cant see what you are writing, i dont understand where you derive them from..

    • 2 years ago
  15. henpen Group Title
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    You could try Taylor expansion for the thing

    • 2 years ago
  16. henpen Group Title
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    Oh, OK\[\large log_k a/log_kb=\log_k(a-b)\]\[\large log_k a \cdot log_kb=\log_k(a+b)\]\[\large log_a b=\log_{a^c}b^c\] Are these clear?

    • 2 years ago
  17. bronzegoddess Group Title
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    the first two.. last one i have never seen..

    • 2 years ago
  18. henpen Group Title
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    It's fairly obvious if you think about it|dw:1352238416055:dw|

    • 2 years ago
  19. bronzegoddess Group Title
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    ok

    • 2 years ago
  20. bronzegoddess Group Title
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    i'll ask my prof tmw :)

    • 2 years ago
  21. henpen Group Title
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    I don't think\[x=(x-a)^{lna}\] is correct, just insert a=e and you get \[x=x-e \] Probably I've messed up somewhere

    • 2 years ago
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is replying to Can someone tell me what button the professor is hitting...

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