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anonymous

  • 5 years ago

logarithms anyone?

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  1. bahrom7893
    • 5 years ago
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    i gotta revise those too, so what are the questions?

  2. anonymous
    • 5 years ago
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    how do i get the inverse y=logsub3x

  3. anonymous
    • 5 years ago
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    so its y=(little3at the bottom)x

  4. bahrom7893
    • 5 years ago
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    k

  5. bahrom7893
    • 5 years ago
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    \[y=\log_{3}x \] \[3^{y}=x\] So x = y^3 is that what u mean by inverse?

  6. anonymous
    • 5 years ago
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    the inverse of the function

  7. anonymous
    • 5 years ago
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    i dont get that either

  8. bahrom7893
    • 5 years ago
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    http://www.andrews.edu/~calkins/math/webtexts/numb17.htm sorry havent done logs in a long time, i only use Ln,s but that is the inverse function

  9. bahrom7893
    • 5 years ago
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    inverse function is: Say: y = f(x) then inverse is: x = F^-1 (y), which basically means solve for X

  10. anonymous
    • 5 years ago
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    ok thank you soo much

  11. bahrom7893
    • 5 years ago
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    hey can u check the answer key.. like gimme any problem with an answer to it

  12. bahrom7893
    • 5 years ago
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    and ill make sure the answer's right

  13. anonymous
    • 5 years ago
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    answer key? like in the back of the book?

  14. anonymous
    • 5 years ago
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    ummm, this is a unit 4 test that was left for hw, so we cant look at the answers in the back of the book T_T

  15. bahrom7893
    • 5 years ago
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    just gimme any similar problems from the book that has an answer to it

  16. bahrom7893
    • 5 years ago
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    like anything like y = log(sub5)x and then it has an answer.. numbers dont matter

  17. anonymous
    • 5 years ago
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    ok, wait a sec

  18. anonymous
    • 5 years ago
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    13. y=logsubX

  19. anonymous
    • 5 years ago
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    answer: f^-19x0=7^x

  20. anonymous
    • 5 years ago
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    *answer: f^-1(x)=7^x

  21. bahrom7893
    • 5 years ago
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    u forgot to add what is under sub

  22. anonymous
    • 5 years ago
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    sorry the second is the real answer

  23. anonymous
    • 5 years ago
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    there is no under sub

  24. bahrom7893
    • 5 years ago
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    Oh its just \[y = \log_{} x\]

  25. bahrom7893
    • 5 years ago
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    ?

  26. anonymous
    • 5 years ago
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    ya, its: y=log(undersub7) X

  27. anonymous
    • 5 years ago
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    answer:f^-1(x)=7^x

  28. bahrom7893
    • 5 years ago
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    Okay that makes sense now

  29. anonymous
    • 5 years ago
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    sorry if i confused you!

  30. bahrom7893
    • 5 years ago
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    So my answer wasn't technically wrong but still it was in wrong notation

  31. bahrom7893
    • 5 years ago
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    workin on the first one

  32. anonymous
    • 5 years ago
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    oh alright

  33. bahrom7893
    • 5 years ago
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    \[y=\log_{3}x ; x = 3^{y} ; f ^{-1} (x) = 3^{x}\]

  34. anonymous
    • 5 years ago
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    i understood everything till the last setp

  35. bahrom7893
    • 5 years ago
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    oh for you to make it into inverse, just replace y with x, say: Sin(x) = y; Sin^-1(y) = x

  36. anonymous
    • 5 years ago
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    so would i have to put it as : \[\prime-1(x)=3^{x}\]

  37. anonymous
    • 5 years ago
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    i meant f^-1(x)=3^x

  38. bahrom7893
    • 5 years ago
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    yes

  39. anonymous
    • 5 years ago
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    hmm, i was pretty sure you could have answered it differently. it cant be: y=3^x

  40. bahrom7893
    • 5 years ago
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    http://www.sosmath.com/algebra/logs/log4/log45/log4513/log4513.html

  41. anonymous
    • 5 years ago
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    oh thanks that seems helpful, can i ask 1 more and see if i get it?

  42. bahrom7893
    • 5 years ago
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    yeah

  43. anonymous
    • 5 years ago
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    f(x)=lnx

  44. anonymous
    • 5 years ago
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    ok, i dont even know what to do here,

  45. anonymous
    • 5 years ago
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    we basically solve for x, right?

  46. bahrom7893
    • 5 years ago
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    yup and this is my favourite stuff! so useful in calculus

  47. bahrom7893
    • 5 years ago
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    k so Lnx is the same as Log(sub e)X, where e is approximately 2.71

  48. anonymous
    • 5 years ago
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    right

  49. anonymous
    • 5 years ago
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    oh,. sorry, how can i add you as a fan?

  50. anonymous
    • 5 years ago
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    oh, nvm, haha

  51. bahrom7893
    • 5 years ago
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    sure lol

  52. anonymous
    • 5 years ago
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    ok, so i get it so far

  53. bahrom7893
    • 5 years ago
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    good and start a new topic. this is getting annoying, my browser keeps scrollin up and down. this place is buggy

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