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anonymous

  • 5 years ago

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  1. anonymous
    • 5 years ago
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    \[\log _{17}17^{y}\]

  2. anonymous
    • 5 years ago
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    And how do I get to the desired answer, I'm more confused as to the method.

  3. amistre64
    • 5 years ago
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    log17(17) = 1 so the answer seems to be "y"

  4. amistre64
    • 5 years ago
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    or to put it another way: 17^? = 17^y ?=y

  5. anonymous
    • 5 years ago
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    Yeah, that's what the book says. Did you get that by putting it in exponential form and then seeing to what multiple the 17 was?

  6. anonymous
    • 5 years ago
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    Y would be 1 in that case?

  7. amistre64
    • 5 years ago
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    y would be any real number in this case...

  8. amistre64
    • 5 years ago
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    there is no = sign to attribute to any specific value

  9. amistre64
    • 5 years ago
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    Do you recall that logs are just exponents?

  10. anonymous
    • 5 years ago
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    it just said "Find:"

  11. anonymous
    • 5 years ago
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    yeah, put in a different form

  12. amistre64
    • 5 years ago
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    B^x = y logB(B^x) = logB(y) x = logB(y)

  13. anonymous
    • 5 years ago
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    meaning, they're exponents reconfigured to find the y correct?

  14. amistre64
    • 5 years ago
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    when we have an "exponent" with an exponent; what can we do with those exponents? multiply them together right? b^2^4 = b^8 right?

  15. amistre64
    • 5 years ago
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    yes

  16. anonymous
    • 5 years ago
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    yes

  17. amistre64
    • 5 years ago
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    so, logB(3^2) is just an "exponent" with an exponent..simply put logB(3^2) = 2 logB(3) does that make sense?

  18. anonymous
    • 5 years ago
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    It makes more sense.

  19. anonymous
    • 5 years ago
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    Still shaky, but I'm getting there.

  20. amistre64
    • 5 years ago
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    good :) cause making less sense is bad lol

  21. anonymous
    • 5 years ago
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    Haahaha

  22. anonymous
    • 5 years ago
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    So I need to learn how to reconfigure my exponents then.

  23. amistre64
    • 5 years ago
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    in a sense, yes :)

  24. anonymous
    • 5 years ago
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    Into logarithmic forms and such. Kinda thinking out loud.

  25. amistre64
    • 5 years ago
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    logB(B) = what? B^? = B ?? B^1 = B right? so, logB(B) = 1

  26. anonymous
    • 5 years ago
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    those b's are bases?

  27. anonymous
    • 5 years ago
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    of exponents?

  28. amistre64
    • 5 years ago
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    take your log now: log17(17^y) turns into: y log17(17) which becomes: y (1) = y

  29. amistre64
    • 5 years ago
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    B = base, yes

  30. anonymous
    • 5 years ago
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    Damn dude, you're a mathemagician

  31. amistre64
    • 5 years ago
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    lol ..... its the rabbit up my sleeve that gets the credit :)

  32. anonymous
    • 5 years ago
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    hi-o

  33. anonymous
    • 5 years ago
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    Alright, well i really appreciate the tutelage.

  34. amistre64
    • 5 years ago
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    no prob...

  35. anonymous
    • 5 years ago
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    i'm sure i'll be back for more haha. peace.

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