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bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
\[a^{x}=\exp(xlna)\]
 2 years ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
and \[\log _{a}=\frac{ lnx }{ lna }\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
For the second, do you mean\[\log_a x=\frac{lnx}{lna}\]?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\large \ln x^{\log_a e} ln a= lnx\] \[\large ( x^{\log_a e} + a)=x\]
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
\[\large x^{\log_a e}=xa\] \[\large x^{\log_a e}=x^{log_x{(xa)}}\] \[\large {\log_a e}={log_x{xa}}\]
 2 years ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
@henpen , yes sorry i forgot the x
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
I'm not quite sure where to go after that
 2 years ago

bronzegoddess Group TitleBest 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?
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352237942749:dw
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
dw:1352237963548:dw
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
Or...dw:1352238104786:dw
 2 years ago

bronzegoddess Group TitleBest 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..
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
You could try Taylor expansion for the thing
 2 years ago

henpen Group TitleBest 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?
 2 years ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
the first two.. last one i have never seen..
 2 years ago

henpen Group TitleBest ResponseYou've already chosen the best response.1
It's fairly obvious if you think about itdw:1352238416055:dw
 2 years ago

bronzegoddess Group TitleBest ResponseYou've already chosen the best response.0
i'll ask my prof tmw :)
 2 years ago

henpen Group TitleBest 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
 2 years ago
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