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iheartfood Group Title

Simplify the expression attached and write it as a single logarithm.. ***still confused on this topic... pls help? :)

  • one year ago
  • one year ago

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  1. iheartfood Group Title
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    • one year ago
  2. iheartfood Group Title
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    answer choices A,B,C,D from top to bottom :)

    • one year ago
  3. tkhunny Group Title
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    Move all the exponents up: \((x+4)^{-3}\) \((x-7)^{2}\) \((x-2)^{5}\) \((x^{2})^{-1}\) Notice how I deliberately left everything outside as ADDITION. This allows us just to bring everything inside without concern for numerator or denominator. We can figure all that out later.

    • one year ago
  4. iheartfood Group Title
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    im confused.. so how can i apply that with my expression?

    • one year ago
  5. tkhunny Group Title
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    Confused with what? \(-3\log(x+4) = \log\left[\left(x+4\right)^{-3}\right]\) Correct?

    • one year ago
  6. iheartfood Group Title
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    im not quite sure what we are working on now.. I'm like uber confused.. are we doing my problem? or an example? :/ sorry I'm a bit confused :(

    • one year ago
  7. tkhunny Group Title
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    Your problem statement begins \(-3\log(x+4)\), doesn't it? Or am I looking at some other picture?

    • one year ago
  8. iheartfood Group Title
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    yup iy does :) so i get log[(x+4)^−3] from that part?

    • one year ago
  9. tkhunny Group Title
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    Very good. Now the (x-7) part?

    • one year ago
  10. iheartfood Group Title
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    2log(x-7) = log(x-7)^2 ??

    • one year ago
  11. tkhunny Group Title
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    Good, now the x-2...

    • one year ago
  12. iheartfood Group Title
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    5log(x-2)=log(x-2)^5 ??

    • one year ago
  13. tkhunny Group Title
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    One more. You're on a roll!

    • one year ago
  14. iheartfood Group Title
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    lol thanks haha :P but i don't get this one... it looks different.. :/ -logx^2...

    • one year ago
  15. iheartfood Group Title
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    but heres my take on it.. log(x^2)^-1 ??

    • one year ago
  16. tkhunny Group Title
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    Take the negative (-1) inside, just like the other coefficients ==> exponents.

    • one year ago
  17. iheartfood Group Title
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    was what i got right?? or no?

    • one year ago
  18. iheartfood Group Title
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    or is it log(x)^-2 ?

    • one year ago
  19. tkhunny Group Title
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    There it is. They are now all connected by addition. Use this guy and bring them all together. log(a) + log(b) = log(a*b)

    • one year ago
  20. iheartfood Group Title
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    sorry my computer crashed... one sec :/

    • one year ago
  21. iheartfood Group Title
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    log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2=answer B right? :)

    • one year ago
  22. jim_thompson5910 Group Title
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    use the rule described above to go from log[(x+4)^−3]+ log(x-2)^5+log(x-7)^2 + log(x)^-2 to log[ (x+4)^−3*(x-2)^5*(x-7)^2*(x)^-2 ]

    • one year ago
  23. iheartfood Group Title
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    kk I'm seeing that :) but how can i simplify it even further?

    • one year ago
  24. jim_thompson5910 Group Title
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    (x+4)^−3 is really 1/[ (x+4)^3 ]

    • one year ago
  25. jim_thompson5910 Group Title
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    same idea applies to (x)^-2

    • one year ago
  26. iheartfood Group Title
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    ok so i get 1/x^-2 ?

    • one year ago
  27. jim_thompson5910 Group Title
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    yep

    • one year ago
  28. jim_thompson5910 Group Title
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    so put that all together

    • one year ago
  29. iheartfood Group Title
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    so i get answer C right?? :)

    • one year ago
  30. iheartfood Group Title
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    sorry i meant answer D :) is it answer D then? :)

    • one year ago
  31. iheartfood Group Title
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    @jim_thompson5910 :)

    • one year ago
  32. jim_thompson5910 Group Title
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    yes it is D

    • one year ago
  33. iheartfood Group Title
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    kk great!!! thx :)

    • one year ago
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