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bronzegoddess
 2 years ago
prove the rule:
bronzegoddess
 2 years ago
prove the rule:

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

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

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

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

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

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

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

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

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

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

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

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

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

henpen
 2 years ago
Best ResponseYou've already chosen the best response.1Oh, 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
 2 years ago
Best ResponseYou've already chosen the best response.0the first two.. last one i have never seen..

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

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

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