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anonymous

  • one year ago

log abc^0.5+log abc =?

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  1. anonymous
    • one year ago
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    options are: 5,9,2,1

  2. xapproachesinfinity
    • one year ago
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    let me see if this what you wrote before we begin \[\log abc^{\frac{1}{2}}+\log abc\]

  3. xapproachesinfinity
    • one year ago
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    is that right?

  4. xapproachesinfinity
    • one year ago
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    is it (abc)^0.5 or just ab(c^0.5) huge difference

  5. xapproachesinfinity
    • one year ago
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    so which is it?

  6. xapproachesinfinity
    • one year ago
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    hello!! you there?

  7. anonymous
    • one year ago
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    yes the first one is ryt

  8. anonymous
    • one year ago
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    will be 1/2log abc +log abc

  9. xapproachesinfinity
    • one year ago
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    so it is (abc)^1/2 yes

  10. anonymous
    • one year ago
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    yes

  11. xapproachesinfinity
    • one year ago
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    okay then \(\log (abc)^{frac{1}{2}}+\log abc= \frac{1}{2} \log abc+\log abc\) \(=\frac{3}{2}\log(abc)\)

  12. xapproachesinfinity
    • one year ago
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    that first line where frac12 supposed to be all to 1/2

  13. xapproachesinfinity
    • one year ago
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    so we get \(=\log(abc)^{\frac{3}{2}}\)

  14. anonymous
    • one year ago
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    yea , next step

  15. xapproachesinfinity
    • one year ago
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    that is all what is it you want to achieve?

  16. xapproachesinfinity
    • one year ago
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    a different way of doing that is \(\log (abc)^{\frac{1}{2}}+\log abc=\log \frac{(abc)^{\frac{1}{2}}}{abc}\)

  17. anonymous
    • one year ago
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    but the options of this ques given are a) 5 b) 9 c) 2 d) 1 which onw wud be the ans?

  18. xapproachesinfinity
    • one year ago
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    i meant \(=log[(abc)^0.5 (abc)]\) not division!!

  19. xapproachesinfinity
    • one year ago
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    \(=\log[(abc)^{0.5} (abc)]\)

  20. xapproachesinfinity
    • one year ago
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    if those are your options then you missing something in the question

  21. anonymous
    • one year ago
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    alright

  22. xapproachesinfinity
    • one year ago
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    can you post the picture of your question

  23. anonymous
    • one year ago
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    that must be the wrong ques. i pick this ques from mformaths.

  24. anonymous
    • one year ago
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    bdw thankew for your help. :)

  25. xapproachesinfinity
    • one year ago
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    that question cannot be any of those number no more simplification is required

  26. anonymous
    • one year ago
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    yep

  27. xapproachesinfinity
    • one year ago
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    all you can do is something like this \(=3/2[\log a+\log b+ \log c]\) which pretty much does not help for anything new

  28. anonymous
    • one year ago
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    ok i have another question cud u help me in that?

  29. xapproachesinfinity
    • one year ago
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    post is in a different post though

  30. anonymous
    • one year ago
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    okz

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