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

• there is abit of mistake in the given ques... both terms contains power 1/2
  its log6(x-3)(x-4)=1
  x^2-7x+12=6
  x^2-7x+6=0
  x=1,6
  Hence 2 real solutions...
• 9 years agoHelpfull: Yes(2) No(1)
• log6 (x^2-6x+9)^1/2 +log6(x^2-4x+16)=1
  =>log6 (x-3)^2 + log6 (x^2-4x+16) = 2
  => (x-3)^2(x^2-4x+16) = 6^2
  => for real solutions
  => (x-3)^2=6^2
  => x-3 = 6 and x-3 = -6
  => x=9,-3
  Hence 2 real solutions
• 9 years agoHelpfull: Yes(1) No(4)
• option b) 2
  was not in option
• 9 years agoHelpfull: Yes(1) No(0)
• i think it will have 5 solns
• 9 years agoHelpfull: Yes(0) No(0)
• d
  
  
  
  log6(x-3)+log6(x^2-4x+16)=1
  log6(x-3)(x^2-4x+16)=1
  (x-3)(x^2-4x+16)=1
  it is not possible to get any solution
  
• 9 years agoHelpfull: Yes(0) No(0)