text: # 1 + 2log 3 - log x = log(x+1)What is x?
  
  giorgiana1976 | College Teacher | (Level 3) Valedictorian
  
  Posted on
  
  First, we'll impose the constraints of existence of logarithms:
  
  x>0
  
  and
  
  x+1>0
  
  x>-1
  
  The common interval of admissible values for x is (0 , +inf.).
  
  We'll re-write the term  2log 3 using the power rule of logarithms:
  
  2log 3 = log 3^2
  
  2log 3 = log 9
  
  We'll re-write the equation, moving all terms that contain the variable x to one side:
  
  1 + log 9 = log(x+1) + log x
  
  Now, we'll solve the equation, applying th product rule of the logarithms:
  
  log9 + log 10 = lg(x+1) + lgx
  
  log 9*10 = log[x*(x+1)]
  
  Because the bases of logarithms are matching, we'll use the one to one property:
  
  x*(x+1)=9*10
  
  We'll remove the brackets and we'll move all terms to one side:
  
  x^2 +x -90=0