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problem on surds..

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  • ravindra ahakey
    Given 3^x=5^y=(75) ^z. then find xy/2x+y..? options-1. z2. z/23. 2z4. 1 solution:-  3^x=5^y=(75) ^z=k   x=log(k) / log 3   ; y=log(k) / log 5  ; z=log(k)
    Message 1 of 5 , Mar 5, 2010
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      Given 3^x=5^y=(75) ^z.


      then find xy/2x+y..?

      options-
      1. z
      2. z/2
      3. 2z
      4. 1

      solution:- 
      3^x=5^y=(75) ^z=k
        x=log(k) / log 3   ; y=log(k) / log 5
       ; z=log(k) / log75
      xy/2x+y = {log(k)/log3} {log(k)/log3}
                        """""""""""""""""""""""""""""""""'
                       2[log(k)/log3] + [log(k)/log5]
      log(k) is canceled from neu and den and take lcm and solve it  to
      =   log(k) / 2log5 + log3  = log(k)/log [(5^2)*3]
      = log(k)/log 75 = z

      answer is 1. z


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    • santanu pradhan
      3^x=5^y=75^z xy/(2x+y)? Answer of this Q as fallows log(3^x)=log(75^z) x=zlog75/log3 log(5^y)=log(75^z) y=zlog75/log5
      Message 2 of 5 , Mar 7, 2010
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        3^x=5^y=75^z
        xy/(2x+y)?
        Answer of this Q as fallows
        log(3^x)=log(75^z)
        x=zlog75/log3
        log(5^y)=log(75^z)
        y=zlog75/log5
         
        ((zlog75/log3)*(zlog75/log5))/((2*zlog75/log3)+zlog75/log5)=z

        so answer=z;

        santanu


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