Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

bronzegoddess Group Title

prove the rule:

  • one year ago
  • one year ago

  • This Question is Closed
  1. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  2. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  3. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  4. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  5. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  6. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  7. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\large x^{\log_a e}=x-a\] \[\large x^{\log_a e}=x^{log_x{(x-a)}}\] \[\large {\log_a e}={log_x{x-a}}\]

    • one year ago
  8. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    @henpen , yes sorry i forgot the x

    • one year ago
  9. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    I'm not quite sure where to go after that

    • one year ago
  10. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  11. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1352237942749:dw|

    • one year ago
  12. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1352237963548:dw|

    • one year ago
  13. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    Or...|dw:1352238104786:dw|

    • one year ago
  14. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  15. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    You could try Taylor expansion for the thing

    • one year ago
  16. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
  17. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

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

    • one year ago
  18. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

    It's fairly obvious if you think about it|dw:1352238416055:dw|

    • one year ago
  19. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    ok

    • one year ago
  20. bronzegoddess Group Title
    Best Response
    You've already chosen the best response.
    Medals 0

    i'll ask my prof tmw :)

    • one year ago
  21. henpen Group Title
    Best Response
    You've already chosen the best response.
    Medals 1

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

    • one year ago
    • Attachments:

See more questions >>>

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
  • 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.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.