bronzegoddess
prove the rule:



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bronzegoddess
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\[a^{x}=\exp(xlna)\]

bronzegoddess
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and \[\log _{a}=\frac{ lnx }{ lna }\]

henpen
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\[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]

henpen
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For the second, do you mean\[\log_a x=\frac{lnx}{lna}\]?

henpen
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\[\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}\]

henpen
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\[\large \ln x^{\log_a e} ln a= lnx\]
\[\large ( x^{\log_a e} + a)=x\]

henpen
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\[\large x^{\log_a e}=xa\]
\[\large x^{\log_a e}=x^{log_x{(xa)}}\]
\[\large {\log_a e}={log_x{xa}}\]

bronzegoddess
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@henpen , yes sorry i forgot the x

henpen
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I'm not quite sure where to go after that

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

henpen
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dw:1352237942749:dw

henpen
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dw:1352237963548:dw

henpen
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Or...dw:1352238104786:dw

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

henpen
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You could try Taylor expansion for the thing

henpen
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Oh, OK\[\large log_k a/log_kb=\log_k(ab)\]\[\large log_k a \cdot log_kb=\log_k(a+b)\]\[\large log_a b=\log_{a^c}b^c\]
Are these clear?

bronzegoddess
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the first two.. last one i have never seen..

henpen
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It's fairly obvious if you think about itdw:1352238416055:dw

bronzegoddess
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ok

bronzegoddess
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i'll ask my prof tmw :)

henpen
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I don't think\[x=(xa)^{lna}\] is correct, just insert a=e and you get \[x=xe \] Probably I've messed up somewhere