how many solution for x....
1/2(log6(x^2-8x+16)+1/2(log6(x^2-9x+81)=1

a)0
b)1
c)3
d)4

• >>log6((x-4)(x-9))=1.....(i)
  >>log6((x-4)(x-9))=log6(6)
  >>(x-4)(x-9)=6
  by solving..x=3,10
  eqn (i) is true for x=10, not for x=3
  because, log(ab)=log(a)+log(b)
  so, only 1 solution for x=10.
  ans-b
• 10 years agoHelpfull: Yes(23) No(4)
• and
  in place of 9x ,it was 18x
• 10 years agoHelpfull: Yes(12) No(1)
• 1/2(log6(x^2-8x+16)+1/2(log6(x^2-9x+81)=1
  1/2(log6(x^2-8x+16)+(log6(x^2-9x+81)))=1 , takng 1/2 common
  log6(x^2-8x+16)+log6(x^2-9x+81)=2
  log6((x^2-8x+16)(x^2-9x+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
  (x-4)(x-9)=6
  x^2-13x+36 = 6
  x = 3,10
  so, there are two solns....
  
  
  
• 10 years agoHelpfull: Yes(10) No(19)
• in place of 9x ,it was 18x.
  by solving both the eq,we will get
  
  log6((x-4)(x-9))=1
  value of x will be 3
  
  it will have only 1 solution
• 10 years agoHelpfull: Yes(8) No(3)
• 1/2(Log6(x^2-8x+16))+1/2(Log6(X^2-9x+81))=1
  1/2(Log6(x-4)^2)+1/2(Log6(x-9)^2)=1
  log6((x-4)^2*(x-9)^2))=2log6(6)
  so,
  ((x-4)^2)((x-9)^2)=6^2
  either (x-4)^2=36 or (x-9)^2=36
  if:(x-4)^2=36
  x-4=+6 or x-4=-6
  so, x=10 or x=-2
  if:(x-9)^2=36
  x-9=+6 or x-9=-6
  so, x=15 or x= 3
  number of solution is:4
  
• 10 years agoHelpfull: Yes(6) No(10)
• log rule was (log a + log b) = log(a*b)
• 10 years agoHelpfull: Yes(3) No(0)
• if u hv sm common sense thn don't wrt wrng qstns
  
• 10 years agoHelpfull: Yes(1) No(2)
• b,loga+logb=logab
• 10 years agoHelpfull: Yes(0) No(0)
• there is no solution for this q . so ans is a) zero
• 9 years agoHelpfull: Yes(0) No(1)
• Ans is b.
  It satisfy for only 10 not for 3.
  As, log6((x-4)(x-9))=log6(6)
  Put x = 10 in above equation
  log 6 base 6 = log 6 base 6
  Hence proved. ;)
• 7 years agoHelpfull: Yes(0) No(1)