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Point on a unit circle

  1. Dec 2, 2016 #1
    1. The problem statement, all variables and given/known data
    the point (log a, log b) exists on the unit circle. find the value of axb. round to the nearest thousandths.

    2. Relevant equations
    x2 + y2 = 1

    3. The attempt at a solution

    x2+y2 = 1
    loga2+logb2 =1
    2loga+2logb = 1
    2(loga+logb) = 1
    loga + log b = 0.5
    logb = 0.5−loga
    now i try and subsitute logb in
    loga2+(0.5−loga)2 = 1

    when i did this it wouldnt work after the last step. so this is what i tried next

    (loga)2 + (log b)2 = 1
    loga = √1-(logb)2

    then i did

    (√1-(logb)2)2 + (logb)2 = 1
    1-(logb)2 = 1-(log b)4
    -(logb)2 + (logb)4 = 0
    factored out (logb)2 so i got
    1 = logb
    and 101 = b so b = 10 but im not sure if thats right either because it seems iffy to me
    i think i was on the right track with the first one but idk can someone help please
     
  2. jcsd
  3. Dec 2, 2016 #2

    haruspex

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    That makes no sense. There are infinitely many solutions.
    E.g. a=1, b=e gives e;
    ab=1 gives 1;
    a=b gives e√2.
    Please check you have stated the question exactly.
     
  4. Dec 2, 2016 #3
    nope i just checked the question and what i said is exactly whats written. it says: the point (log a, log b) exists on the unit circle. find the value of a times b. round to the nearest thousandth
     
  5. Dec 3, 2016 #4

    haruspex

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    I just noticed you were assuming log base 10, whereas I assumed natural logs, but my answer applies whatever the base. It is more obvious if we substitute x=log a, y= log b, v=log (ab). Let the base be c. ab=cxcy=cx+y, so v=x+y. So it is the same as asking for the value of x+y given that x2+y2=1.
     
  6. Dec 3, 2016 #5

    Mark44

    Staff: Mentor

    No.
    The second equation is ##(\log(a))^2 + (\log(b))^2 = 1##. This is not the same as ##\log(a^2) + \log(b)^2 = 1##.
     
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