If x1/p = y1/q = z1/r and y2 = xz, what is the value of p+r?

• let x^(1/p) = y^(1/q) = z^(1/r) = k
  therefore, x = k^p
  y = k^q
  z = k^r
  now,
  y^2 = xz
  => k^2q = k^p.k^r
  => 2q=p+r......ans
• 10 years agoHelpfull: Yes(43) No(3)
• according to question, x^(q/p)=y and also z^(q/r)=y..so put this in LHS
  lets come to RHS ...we can say that x^(r/p) =z. put this in RHS....
  now u will get x^(q/p).z^(q/r)=x.z ...in place of z put above expression containing z
  so x^(q/p).x^(q/p)=x.x^(r/p)..( do it carefully)
  equate powers:- 2q/p=1+r/p
  u will get 2q=p+r
  
  
  
• 10 years agoHelpfull: Yes(4) No(0)
• x1/p1=y1/q1==>> x1/y1=p/q==>>p=x1, q=y1
  similarly r=z1
  p+r=x1+z1..
  this is the answer as there r no values given
• 10 years agoHelpfull: Yes(1) No(3)
• 2q
  by taking log of the equations given,it becomes
  2q=p+r
• 7 years agoHelpfull: Yes(1) No(0)
• let x^(1/p) = y^(1/q) = z^(1/r) = k
  therefore, x = k^p;
  y = k^q;
  z = k^r;
  now,
  y^2 = xz;
  => k^2q = k^p.k^r;
  => 2q=p+r ans
  Priya "Use ';' at the end of the statement "
• 8 years agoHelpfull: Yes(0) No(0)