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anonymous

  • one year ago

log(p+q)(p-q) = -1 log(p+q)log(p^2-q^2) = ? option: a)2 b)1 c) -1 d) 0

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  1. anonymous
    • one year ago
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    cud anybody give me the full solution. ?

  2. anonymous
    • one year ago
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    umm I don't know if this is correct: \[\log(p+q)\log(p^2-q^2) =?\] \[\log(p+q)\log(p-q) = -1\] so first equation is turned into: \[\log(p+q)\log((p+q)(p-q)) = ?\]

  3. anonymous
    • one year ago
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    since \[\log(p+q)(p-q) = -1\] so \[\log(p+q) -1 =0\]

  4. welshfella
    • one year ago
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    log(p^2-q^2) = log(p+q)(p-q) = -1 but i cant see a way to find what log(p+q) is.

  5. anonymous
    • one year ago
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    @welshfella that's in my last reply

  6. anonymous
    • one year ago
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    log(p+q) = 1

  7. anonymous
    • one year ago
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    so since log(p+q) = 1 it will be: \[\log(p+q)\log(p^2-q^2) = 0\]

  8. welshfella
    • one year ago
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    i cant see your logic there - I'm not saying you are wrong

  9. anonymous
    • one year ago
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    I mean to say that since log(p^2-q^2) = -1 log(p+q)log(p^2-q^2) = will be log(p+q) -1 = 0 so log(p+q) = 1

  10. welshfella
    • one year ago
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    if log ( p + q) = 1 log(p+q)log(p^-q^2) = 1 * -1 = -1 not 0

  11. anonymous
    • one year ago
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    oh yes my bad I was subtracting then yes it would be -1

  12. freckles
    • one year ago
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    @nopen it looked like you said: \[\log(a \cdot b)=\log(a) \cdot \log(b) \text{ thisn't a log property though }\]

  13. hartnn
    • one year ago
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    can the asker post the screenshot of the question?

  14. anonymous
    • one year ago
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    but isn't it log(a*b) = log(a) + log(b)?

  15. freckles
    • one year ago
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    yes that is true

  16. freckles
    • one year ago
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    that is the product rule for log

  17. welshfella
    • one year ago
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    I think the answer is zero let p+q = 1 and p-q = 0.1 then log(p+ q)(p - q) = log 0.1 = -1 which fits in with log(p+q) + log(p-q) = 0 + (-1) = -1 because the log of 1 is 0 so when we multiply by log(p+q) we are multiplying by 0 so the log(p+q)log(p^2 - q^2) = 0*-1 = 0

  18. welshfella
    • one year ago
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    yeah its d

  19. freckles
    • one year ago
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    well I don't think this can be answered with the info given here is a counterexample to not being zero for all p and q that satisfy the condition log(p^2-q^2)=-1 \[\text{ Let } p+q=10^{-2} \text{ and } p-q=10 \\ \text{ so } p^2-q^2=10^{-1} \\ \text{ so } \log((p+q)(p-q))=-1 \\ \text{ but } \\ \log(p+q)\log(p^2-q^2)=(-2)(-1)=2\]

  20. welshfella
    • one year ago
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    yes - that is correct - we need more info to get a definitive answer.

  21. freckles
    • one year ago
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    a picture would be nice as suggested by @hartnn

  22. welshfella
    • one year ago
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    yes

  23. anonymous
    • one year ago
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    thank you @nopen

  24. anonymous
    • one year ago
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    but we cant assume log(p+q)log(p+q)(p-q)=0, as this condition is not given. if log(p+q)(-1)=? that too is in multiplication.

  25. anonymous
    • one year ago
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  26. freckles
    • one year ago
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    \[\text{ Let } p+q=10^{A} \text{ then } p-q=10^{-1-A} \\ \text{ Since we need } \log[(p+q)(p-q)]=-1 \\ \text{ Verify our let part: } \\ \log[(10^{A})(10^{-1-A})]=\log[10^{A+(-1-A)}[=\log(10^{-1})=-1 \log(10)=-1(1)=-1 \\ \text{ Verified! } \\ \log(p+q)=\log(10^{A})=A \log(10)=A(1)=A \\ \log(p+q)\log(p^2-q^2)=A(-1)=-A\] A can be any real number

  27. anonymous
    • one year ago
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    and what wud be the answer in this case , 1 or 2?

  28. freckles
    • one year ago
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    no way to tell

  29. freckles
    • one year ago
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    Not enough info is given to apply one of the choices

  30. freckles
    • one year ago
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    I wonder if they mean (p+q) to be the base ... if so... the problem really is: \[\log_{p+q}(p-q)=-1\] so we have \[(p+q)^{-1}=p-q\] and we are trying to find: \[\log_{p+q}((p-q)(p+q)\] using product rule we have \[\log_{p+q}(p-q)+\log_{p+q}(p+q)\] \[\log_{p+q}((p+q)^{-1})+1\] using power rule (-1)+1=0 now this is all assuming p+q is in the interval (0,1) union (1,infty)

  31. anonymous
    • one year ago
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    waooo , Great ... thanku @freckles

  32. freckles
    • one year ago
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    no sorry the question doesn't make sense lol because they have more than one log in the thingy we are supposedly suppose to find

  33. anonymous
    • one year ago
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    but it seems good

  34. anonymous
    • one year ago
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    we are not given with any base , so we have to identify by ourself that the base is given or not

  35. freckles
    • one year ago
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    there is seriously not enough information to answer this question the way it is written I have been trying to find other ways to possibly interpret the problem but i do not see a way to interpret the problem really that will lead to one of the answers but anyway I interpret it like the above I have leave something out

  36. anonymous
    • one year ago
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    that is not ryt

  37. freckles
    • one year ago
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    what are you talking about?

  38. freckles
    • one year ago
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    the answer?

  39. freckles
    • one year ago
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    I didn't give an answer to your problem

  40. anonymous
    • one year ago
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    the above way that you have used to solve the question

  41. anonymous
    • one year ago
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    taking base p+q

  42. freckles
    • one year ago
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    I made up a question that could have a answer that is one of your choices because the question you have could give us any of the choices (being that there is not enough information)

  43. anonymous
    • one year ago
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    fine

  44. anonymous
    • one year ago
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    answer is that, the question doesnt make any sense,. lol

  45. hartnn
    • one year ago
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    lol if p+q is the base, then this is a really simple problem, log_(p+q) of p-q = -1 log_(p+q) (p^2-q^2) = log_(p+q) of (p-q)(p+q) =log_(p+q) of p-q + log_(p+q) of p+q = -1+1=0

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