A community for students.
Here's the question you clicked on:
 0 viewing
bronzegoddess
 3 years ago
prove the rule:
bronzegoddess
 3 years ago
prove the rule:

This Question is Closed

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

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

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

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

henpen
 3 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
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0@henpen , yes sorry i forgot the x

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

bronzegoddess
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1Or...dw:1352238104786:dw

bronzegoddess
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1You could try Taylor expansion for the thing

henpen
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.0the first two.. last one i have never seen..

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

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

henpen
 3 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
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.