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Math Help - logarithms help

  1. #1
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    logarithms help

    prove that if  a^{x} = b^{y} = (ab)^{xy} than  x+y =1

    A not really sure how to start,

    what I have done so far;

     a^{x} = b^{y}

     xlog_{a} a = ylog_{a} b

     x =ylog_{a} b

    is this correct

    where do I go from here?






     ab^{xy}

     xylog_{a} (ab)

     xlog_{a} a + ylog_{a} b

    is this also correct ?
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  2. #2
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    Hello,

    Quote Originally Posted by Tweety View Post
    prove that if  a^{x} = b^{y} = (ab)^{xy} than  x+y =1

    A not really sure how to start,

    what I have done so far;

     a^{x} = b^{y}

     xlog_{a} a = ylog_{a} b

     x =ylog_{a} b

    is this correct
    this is correct (*)

    where do I go from here?

     ab^{xy}

     xylog_{a} (ab)

     xlog_{a} a + ylog_{a} b

    is this also correct ?
    This is not correct (*), xy \log_a(ab)=xy \log_a(a)+xy\log_a(b)=xy+xy \log_a (b)


    (*) if a and b are negative, there will be some problems for defining the logarithm...


    Anyway, here is what I did :
    - if a,b=0, then the equality is always true, for any x and y. So it is true to say that x+y=1
    - if a,b=1, same as above
    - if a,b \neq 0 or 1 :
    \begin{aligned}<br />
a^x &=(ab)^{xy} \\<br />
&=a^{xy}b^{xy} \\<br />
&=a^{xy} (b^y)^x \\<br />
&=a^{xy} (a^x)^x \\<br />
&=a^{xy} a^{x^2} \\<br />
&=a^{x^2+xy} \end{aligned}

    1=a^{x^2+xy-x}
    this is true iff x^2+xy-x=0, that is to say x(x+y-1)=0
    similarly for b, we get y(x+y-1)=0

    if x=0,
    - if x+y-1=0, then y=1 and we have x+y=1
    - if y=0, then there is a problem here...

    if x+y-1=0, everything is ok.



    I wonder if you stated all the conditions over a,b,x and y...
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