If log(P)=(1/2) log(q)=(1/3)log(r), then which of the following option is true..??
1- p2=q3r2
2- q2=pr
3- q2=r3p
4- r=p2q2

• log p=1/2logq, q=p2
  logq=1/3logr, r=q3
  r=q.q2
  =p2q2
• 9 years agoHelpfull: Yes(10) No(3)
• answer:option 2
  log(p)=1/2log(q) ----> q=p^2
  log(p)=1/3log(r) -----> r=p^3
  let assume p=2,q=4,r=8 which satisfies the above relations,then verify the options
  q^2=4^2=16
  p*r=2*8=16
• 9 years agoHelpfull: Yes(5) No(0)
• option 2
  log p=k p=e pow k q=e pow k pow 2 r = e pow k pow 3
  so option 2 satisfies
• 9 years agoHelpfull: Yes(2) No(0)
• from the property of log
  p=q^1/2=r^1/3
  so, p^2=q
  p^3=r
  now, q^2=p^4
  =P^3.P
  q =r.p
• 8 years agoHelpfull: Yes(1) No(0)
• log p = 1/2*log q
  log p = log q^1/2
  p = q^1/2
  
  log q^1/2=log r^1/3
  q^1/2 = r^1/3
  
  now p=q^1/2
  p *p = q^1/2 *p
  p^2 = q^1/2 *r^1/3
  ans:option 1
• 9 years agoHelpfull: Yes(0) No(4)
• p=1/2
  q=1/3
  so r=1/4
  apply on option..
  in 2nd option q*2=p*r
  2/3=(1/2)*(1/3)
  =1/4
  
• 9 years agoHelpfull: Yes(0) No(2)
• log p=1/2 log q
  p^2=q ----------------eq1
  lop p=1/3 log r
  p^3= r ----------------------eq2
  relating eq1 and eq2
  r=p^3=p^2*p (p^2=q)
  r=pq
  q=r/p
  squaring both sides
  q^2=r^2/p^2
  q^2=(r^2/p^3)*p
  
  q^2=(r^2/r)*p ----------------....(p=r^3)
  q^2=rp ans(2)
  solved by shambhu Mahato
• 8 years agoHelpfull: Yes(0) No(0)
• p6=q3=r2
  q=p2
  r=p3
  so, q2=p4
  pr=p4
  so,q2=pr
• 8 years agoHelpfull: Yes(0) No(0)
• log p =log q^1/2 => p=q^1/2
  
  log q^1/2 = log r^1/3 => q^1/2=r^1/3
  
  by considering option b
  i.e
  q^2=pr
  
  because
  
  q^1/2 *q^1/2 *q^1/2 *q^1/2 = p * p*p * p
  = p * r^1/3 * r^1/3 *r^1/3
  = p * r^3/3 [since when bases are equal powers should be added]
  = P * r^1
  = pr
• 8 years agoHelpfull: Yes(0) No(0)
• log(p)=1/2log(q)=1/3log(r)=k
  p=e^k;
  q=e^2k;
  r=e^3k ; q^2=p.r;( e^2k)^2=e^k.e^3k=>e^4k
• 8 years agoHelpfull: Yes(0) No(0)
• log(p)=1/2log(q)=1/3log(r)
  p=q^1/2=r^1/3
  p^2=q
  p^3=r
  q^6=r^2
  then we get q^2=pr
• 6 years agoHelpfull: Yes(0) No(0)