1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
How many real solutions of x exist?

• 
  1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
  
  => 1/2[ log base 6 of (x^2-8x+16) + log base 6 of(x^2-18x+81)]= 1
  
  => log base 6 of (x-4)^2 + log base 6 of (x-9)^2 = 2
  
  => log base 6 of [(x-4)^2*(x-9)^2] = log base 6 of 36
  
  => (x-4)^2 * (x-9)^2 = 36
  
  => (x-4)*(x-9)=6 0r (x-4)*(x-9)= -6
  
  => x^2-13x+30=0 0r x^2-13x+42=0
  
  => (x-3)(x-10)= 0 or (x-6)(x-7)= 0
  
  => x= 3,10,6,7
  
  all 4 values satisfies the given eqn,
  
  4 real solns
  
• 10 years agoHelpfull: Yes(41) No(5)
• 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
  
  => 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))= 1
  
  => log base 6 of (x^2-8x+16)^(1/2) + log base 6 of(x^2-18x+81)^(1/2)= 1
  
  => log base 6 of {(x-4)^2}^(1/2) + log base 6 of {(x-9)^2}^(1/2) = 1
  
  => log base 6 of (x-4) + log base 6 of (x-9) = 1
  
  => log base 6 of {(x-4)*(x-9)} = log base 6 of 6
  
  => (x-4)*(x-9)= 6
  => x^2-13x+30 = 0
  => (x-3)(x-10)= 0
  => x = 3,10
  
  both x=3 & x=10 satisfies given eqn
  
  no. of real solutions of x = 2
• 10 years agoHelpfull: Yes(26) No(13)
• only one real solution exist x=10.
  
  =>log base 6 of {(x-4)(x-9)}=1
  =>{(x-4)(x-9)}>0
  Because x>9 x=3 is not the solution of this equation.
• 10 years agoHelpfull: Yes(6) No(11)
• 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
  
  => 1/2[ log base 6 of (x^2-8x+16) + log base 6 of(x^2-18x+81)]= 1
  
  => log base 6 of (x-4)^2 + log base 6 of (x-9)^2 = 2
  
  => log base 6 of [(x-4)^2*(x-9)^2] = log base 6 of 36
  
  => (x-4)^2 * (x-9)^2 = 36
  
  => (x-4)*(x-9)=6 0r (x-4)*(x-9)= -6
  
  => x^2-13x+30=0 0r x^2-13x+42=0
  
  => (x-3)(x-10)= 0 or (x-6)(x-7)= 0
  so x= 3,10,6,7
  
  and only the 4 solution is here
• 10 years agoHelpfull: Yes(5) No(4)
• 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81)) = 1
  
  => 1/2(log base 6 of (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))= 1
  
  => log base 6 of (x^2-8x+16)^(1/2) + log base 6 of(x^2-18x+81)^(1/2)= 1
  
  => log base 6 of {(x-4)^2}^(1/2) + log base 6 of {(x-9)^2}^(1/2) = 1
  
  => log base 6 of (x-4) + log base 6 of (x-9) = 1
  
  => log base 6 of {(x-4)*(x-9)} = log base 6 of 6
  
  => (x-4)*(x-9)= 6
  => x^2-13x+30 = 0
  => (x-3)(x-10)= 0
  => x = 3,10
  Now
  =>log base 6 of {(x-4)(x-9)}=1
  =>{(x-4)(x-9)}>0
  for x=10
  (x-4)(x-9)=6>0
  for x=3
  (x-4)(x-9)=-1*-6=6>0
  So,
  no. of real solutions of x is 2
  
  
• 10 years agoHelpfull: Yes(4) No(3)
• 1/2(log base 6 0f (x^2-8x+16)) + 1/2(log base 6 of(x^2-18x+81))=1
  
  => 1/2 ( log base 6 of (x-4)^2 + log base 6 0f (x-9)^2 )=1
  
  => log base 6 0f ((x-4)^2*(x-9)^2)=2
  
  => log base 6 of ( (x-4)^2*(x-9)^2)= log base 6 of 36
  
  => (x-4)^2 * (x-9)^2 = 36
  
  So, now possible L.H.S. will be - 1 * 36 and 4 * 9 ( bec. squares
  multiplication is 36 )
  So, x can be x=3 , x= 6 and x=10 . So, 3 real values possible.
• 10 years agoHelpfull: Yes(1) No(5)
• Ans. is 4
  log base 6^2 of (x^2-8*x+16)+log base 6^2 of (x^2-18*x+81)=1
  log base 36 of (x^2-8*x+16)(x^2-18*x+81)=1
  (x^2-8*x+16)(x^2-18*x+81)=36
  (x-4)^2*(x-9)^2=6^2
  (x-4)*(x-9)=-6 or (x-4)*(x-9)=+6
  here x=3,10 here x=6,7
  so, totally 4
• 10 years agoHelpfull: Yes(1) No(0)
• The problem here is, for guys who are saying the solutions are 2.The options were 0,1,3,4
• 10 years agoHelpfull: Yes(0) No(0)
• x>5 and x>-8
  
• 10 years agoHelpfull: Yes(0) No(0)
• what is exact answer ?
  
• 8 years agoHelpfull: Yes(0) No(0)