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bronzegoddess

  • 2 years ago

prove the rule:

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  1. bronzegoddess
    • 2 years ago
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    \[a^{x}=\exp(xlna)\]

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

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

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

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

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

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

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

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

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

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

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

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

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

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

  16. henpen
    • 2 years ago
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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?

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

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

  19. bronzegoddess
    • 2 years ago
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    ok

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

  21. henpen
    • 2 years ago
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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

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