log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1
Find the no of real solution.
a.)4 b.)2 c.)3 d.)0
note:-here log of base 6

• by solving,we get
  log6((x-3)(x-4))=1
  when x=6&x=1 it will satisfy the equation
  hence it will have 2 real solution
• 10 years agoHelpfull: Yes(21) No(3)
• from above eqn: log6((x-3)^2(x-4)^2)=2
  log6{(x-3)(x-4)}^2 }=2
  {(x-3)(x-4)}^2 }=36
  (x-3)(x-4)=+6
  x^2-7x+6=0
  gives x=1 nd x=6
  and (x-3)(x-4)=-6
  x^2-7x+18=0
  discrimant
• 10 years agoHelpfull: Yes(4) No(3)
• rest part
  above eqn give imgnry root..left them
  nw we have x=1 and x=6
  it also nt satisfied orignal eqn
  log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1
  ans=d
• 10 years agoHelpfull: Yes(2) No(3)
• By Solving this we get,
  log((x-3)(x-4))=log6
  so, (x-3)(x-4)=6
  
  above is clearly showing that x will have 2 solution
• 10 years agoHelpfull: Yes(1) No(1)
• in place of 9x ,it was 18x.
  by solving both the eq,we will get
  1/2(log6(x^2-8x+16)+1/2(log6(x^2-18x+81)=1
  1/2(log6(x^2-8x+16)+(log6(x^2-18x+81)))=1 , takng 1/2 common
  log6(x^2-8x+16)+log6(x^2-18x+81)=2
  log6((x^2-8x+16)(x^2-18x+81))=2 , acc to log rule log a + log b = log(a+b)
  log6((x-4)^2(x-9)^2)=2
  log6((x-4)(x-9))^2=2
  2log6((x-4)(x-9))=2
  log6((x-4)(x-9))=1
  >>log6((x-4)(x-9))=1.....(i)
  >>log6((x-4)(x-9))=log6(6)
  >>(x-4)(x-9)=6
  by solving..x=3,10 ans-b
  it will have only 2 solution
  
• 10 years agoHelpfull: Yes(1) No(3)
• Ans:b)2
  Sol: log6(x-3)(x-4)=1
  after solve ,we get
  36x^2-252x-422=0
  D=252^2-4*36*422>0
  so D>0
  hence it has two 2 real solution
• 9 years agoHelpfull: Yes(0) No(0)