IF LOG2,LOG(2^X+1),LOG(2^X+3) ARE IN A.P. THEN WHAT IS THE VALUE OF X

• (2^x+1)^2=2(2^x+3)
  >2^2x+1=6
  >2^2x=5
  >x=0.5 log5/log2
• 11 years agoHelpfull: Yes(33) No(2)
• log(2^x+1)-log(2)=log(2^x+3)-log(2^x+1)
  gives x=2
• 11 years agoHelpfull: Yes(20) No(15)
• If it is 2^(x+1) and 2^(x+3),then x=2 is the solution.
  And if it is 2^x + 1 and 2^x + 3, then the solution is x=log5/log4.
• 11 years agoHelpfull: Yes(7) No(3)
• answer shall be (log 5)/2*(log 2)
• 11 years agoHelpfull: Yes(1) No(0)
• 2log2=log(2^x+1)+log(2^x+3)
  log(2^2)=log((2^x+1)(2^x+3))
  2^2 = 2^(X+1+X+3)
  2=2x+4
  x=-1
• 11 years agoHelpfull: Yes(0) No(5)
• log .5 / log 2
• 11 years agoHelpfull: Yes(0) No(1)
• please anyone explain clearly i am not getting
• 11 years agoHelpfull: Yes(0) No(1)
• log (2^ x+1) - log 2= log (2^x+3) - log (2^ x+1)
  apply log rules
  (2^ x+1)/2 = (2^x+3)/(2^x+1)
  2^2x+2 = 2^x+4
  2x+2 = x+4
  x=2
• 9 years agoHelpfull: Yes(0) No(0)