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bronzegoddessBest ResponseYou've already chosen the best response.0
\[a^{x}=\exp(xlna)\]
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
and \[\log _{a}=\frac{ lnx }{ lna }\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
For the second, do you mean\[\log_a x=\frac{lnx}{lna}\]?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[\large \ln x^{\log_a e} ln a= lnx\] \[\large ( x^{\log_a e} + a)=x\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
\[\large x^{\log_a e}=xa\] \[\large x^{\log_a e}=x^{log_x{(xa)}}\] \[\large {\log_a e}={log_x{xa}}\]
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
@henpen , yes sorry i forgot the x
 one year ago

henpenBest ResponseYou've already chosen the best response.1
I'm not quite sure where to go after that
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
am not sure i follow ur steps.. how did you get all the those terms in the first part?
 one year ago

henpenBest ResponseYou've already chosen the best response.1
Or...dw:1352238104786:dw
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
my problem isnt that i cant see what you are writing, i dont understand where you derive them from..
 one year ago

henpenBest ResponseYou've already chosen the best response.1
You could try Taylor expansion for the thing
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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?
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
the first two.. last one i have never seen..
 one year ago

henpenBest ResponseYou've already chosen the best response.1
It's fairly obvious if you think about itdw:1352238416055:dw
 one year ago

bronzegoddessBest ResponseYou've already chosen the best response.0
i'll ask my prof tmw :)
 one year ago

henpenBest ResponseYou've already chosen the best response.1
I don't think\[x=(xa)^{lna}\] is correct, just insert a=e and you get \[x=xe \] Probably I've messed up somewhere
 one year ago
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