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anonymous

  • one year ago

Simplify 5 log2 k − 8 log2 m + 10 log2 n. 7 log2 (k − m + n) 7 log2 kn over m log2 50 kn over 8 m log2 k to the fifth power n to the tenth power over m to the eighth power

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  1. anonymous
    • one year ago
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    @uri @LeAnn01Xxx @e.mccormick @Lady.Liv1776 @TylerMJ325 @Terrorclaw @Phebe @keshaun455 @JoshKoikkara

  2. anonymous
    • one year ago
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    Hmm... 5*log2 k - 8*log2 m + 10*log2 n log2 (k^5*n^10)/m^8 , I'm not sure...do u get the same?

  3. anonymous
    • one year ago
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    I don't want the answer I want to learn how to do it. I'm really bad at Maths ...

  4. Nnesha
    • one year ago
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    quotient rule\[\large\rm log_b y - \log_b x = \log_b \frac{ x }{ y}\] to condense you can change subtraction to division product rule \[\large\rm log_b x + \log_b y = \log_b( x \times y )\] addition ----> multiplication power rule \[\large\rm log_b x^y = y \log_b x\] remember!

  5. anonymous
    • one year ago
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    @Marcos_RG4L Hmm, I forgot Latex wait.... http://openstudy.com/study#/updates/4ffb5ce1e4b00c7a70c465e9

  6. anonymous
    • one year ago
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    @Nnesha Hmm...Great !!

  7. Nnesha
    • one year ago
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    first apply power property \[\large\rm log_b x^y = y \log_b x\]

  8. anonymous
    • one year ago
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    5 log 2 ?

  9. Nnesha
    • one year ago
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    |dw:1434389999658:dw| 2 is a base number at front of log became exponent of k

  10. Nnesha
    • one year ago
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    \[\huge\rm 5 \log_{2} = k = \log_2 k^5\]

  11. anonymous
    • one year ago
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    Ok Always change ?

  12. Nnesha
    • one year ago
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    \[\huge\rm 5 \log_{2} k = \log_2 k^5\]

  13. Nnesha
    • one year ago
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    yes always so what is \[8 \log_2 m = ???\]

  14. Nnesha
    • one year ago
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    |dw:1434390120641:dw|

  15. anonymous
    • one year ago
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    ok

  16. Nnesha
    • one year ago
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    \[\huge\rm 8 \log_2 m = ???\]try this one

  17. anonymous
    • one year ago
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    m= 2 log 8^k?

  18. Nnesha
    • one year ago
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    there isn't any k in that part

  19. anonymous
    • one year ago
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    Then is only the 8 log 2^ m = 2 log 8^ ?

  20. Nnesha
    • one year ago
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    look at the log property \[\large\rm y \log_b x = log_b x^y\] number at front of log became exponent of x no the base

  21. Nnesha
    • one year ago
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    here is an example \[\huge\rm \color{red}{3} \log_4 \color{blue}{ 7}\]\[\large\rm log_4 \color{blue}{7}^\color{reD}{3}\]

  22. Nnesha
    • one year ago
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    \[\huge\rm \color{reD}{8} \log_2 \color{blue}{m} = ???\]

  23. anonymous
    • one year ago
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    8 log 2 m = log 2 m 8^ ?

  24. Nnesha
    • one year ago
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    yay!!

  25. anonymous
    • one year ago
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    Ok, The colors help me ...

  26. Nnesha
    • one year ago
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    \[\huge\rm log_2 k^5 - \log_2 m^8 + \color{red}{10} \log_2 \color{blue}{n}\] last one apply power rule

  27. anonymous
    • one year ago
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    10 log 2 n = log 2 n 10^ ?

  28. Nnesha
    • one year ago
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    yes right \[\huge\rm log_2 k^5 \color{ReD}{-} \log_2 m^8 \color{blue}{+} \log_2 {n^{10}}\] now look at the addition and subtraction sign sign and look at the properties which one you should apply ?

  29. anonymous
    • one year ago
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    Multiplication?

  30. Nnesha
    • one year ago
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    for which one there are two part negative and plus sign

  31. anonymous
    • one year ago
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    Idk umm. Log 2 m 8^ x log 2 n 10^?

  32. Nnesha
    • one year ago
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    log_2 m^8 is negative so which property is for negative signs ?

  33. Nnesha
    • one year ago
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    you can write it as \[\huge\rm log_2 k^5 + \log_2 n^{10} - \log_2 m^8\]

  34. anonymous
    • one year ago
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    ok

  35. Nnesha
    • one year ago
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    you can write it as \[\huge\rm \color{reD}{log_2 k^5 + \log_2 n^{10}} \color{blue}{ - \log_2 m^8}\] for red part = apply product property for blue part = apply quotient property

  36. anonymous
    • one year ago
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    Thanks for ur help I need to go thank you very much...

  37. Nnesha
    • one year ago
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    alright gO_Od luck!

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