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Proof using logarithm

  1. Nov 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Positive integers a and b, where a < b, satisfy the equation :
    ab = ba

    By first taking logarithms, show that there is only one value of a and b that satisfies the equation and find the value !


    2. Relevant equations
    logarithm


    3. The attempt at a solution
    I know the solution just by guessing, a = 2 and b = 4 but I don't know how to do it...

    [tex]a^b=b^a[/tex]

    [tex]b*log~a=a*log~b[/tex]

    Then I stuck....:frown:

    Thanks
     
  2. jcsd
  3. Nov 12, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    try taking logs to base a or b instead of base 10
     
  4. Nov 12, 2009 #3
    Hi rock.freak667
    I think it will be the same.

    [tex]a^b=b^a[/tex]

    [tex]b*\log_{a}~a=a*\log_{a}b[/tex]

    [tex]b=a*\log_{a}b[/tex]

    Then stuck again...:(

    Thanks
     
  5. Nov 17, 2009 #4

    turin

    User Avatar
    Homework Helper

    Think of only one variable at a time (e.g. b in the last equation of your previous post).
     
  6. Nov 18, 2009 #5
    Since a and b are positive integer,

    we can conclude a,b>0 without generalisation.

    First by taking log base a,

    [tex]

    b=a log_{a}b[/tex]

    [tex]b log_{b}a=a[/tex]

    substituting to initial equation,

    [tex]a^{2}-alog_{a}b=0[/tex]

    [tex]a=0[/tex] or [tex]a=log_{a}b[/tex]

    Therefore, [tex]a=log_{a}b[/tex]

    is the only solution for a since logarithm curve is constantly decreasing.

    [tex]b=a^{a}[/tex]

    since exponent curve is ... leaving this part to you (=
     
    Last edited: Nov 18, 2009
  7. Nov 18, 2009 #6
    Hi turin and icystrike
    sorry I don't understand what you mean. From my last equation :

    [tex]b=a*\log_{a}b[/tex]

    Then, think only b as variable. how to continue? what about a?

    I don't understand this part. To which initial equation do you substitute?

    even though I can reach this part, I still don't know how to continue. I think logarithm curve is constantly increasing, not decreasing. From y = log x, the value of y will increase if x increases.

    exponent curve is also constantly increasing, but from b=aa, how to deduce that a = 2?

    Thanks a lot
     
    Last edited: Nov 18, 2009
  8. Nov 19, 2009 #7

    Mark44

    Staff: Mentor

    I'm with Songoku on this; I'm not following what you are doing. I understand how you got both equations above, but your description is that you are taking the log base a of both sides of the original equation. In your second equation you're taking the log base b of both sides of the original equation.
    Now this doesn't make any sense to me. Where did it come from? This is equivalent to a2 - b = 0 from the first of your two equations above, where you have b = a logab.
    Assuming that the base is larger than 1, any log curve is constantly increasing. IOW, if x1 < x2, then logax1 < logax2.
     
  9. Nov 19, 2009 #8

    turin

    User Avatar
    Homework Helper

    Whoops, I thought I knew how to do the problem, until you pointed out this assumption.
     
  10. Nov 19, 2009 #9

    Mark44

    Staff: Mentor

    So unless icystrike can explain what he's done, we're back at square 1 on this problem.
     
  11. Nov 20, 2009 #10
    Sorry guys i've made a mistake. we are back to square one.
    Now i'm able to solve it.
    But it contradicts the statement that a>b.

    [tex]
    ((lg a)/(lg b))=((lg b)/(lg a))[/tex]
    [tex](lg a)^{2}-(lg b)^{2}=0[/tex]
    [tex](lg a-lg b)(lg a+lg b)=0[/tex]
    Therefore a=b which contradicts a>b
    Thus the only possibility is [tex]a=(1/b)[/tex]
    since a is a interger,
    b must divide 1,
    whereby b=1
    suggest a=1 .
    and resulting a=b.
    Which again leads to a contradiction
     
    Last edited: Nov 20, 2009
  12. Nov 20, 2009 #11

    Mark44

    Staff: Mentor

    How do you get the equation above? We know that ab = ba, so b loga = a logb
    ==> b/a = (log b)/(log a)

    How do you get from this equation to (log a)/(log b) = (log b)/(log a)?
     
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