text: # Solve for x the equation e^2x-6*e^x+5=0
  
  justaguide | College Teacher | (Level 2) Distinguished Educator
  
  Posted on
  
  We have to solve e^2x - 6*e^x + 5 = 0 for x.
  
  Let e^x = y
  
  => y^2 - 6y + 5 = 0
  
  => y^2 - 5y - y + 5 = 0
  
  => y(y - 5) - 1(y - 5) = 0
  
  => (y - 1)(y - 5) = 0
  
  => y = 1 and y =5
  
  e^x = 1 and e^x = 5
  
  => x = 0 and x = ln 5
  
  The solution of the equation are x = 0 and x = ln 5
  
  tonys538 | Student, Undergraduate | (Level 1) Valedictorian
  
  Posted on
  
  The equation e^2x-6*e^x+5=0 has to be solved.
  
  e^2x-6*e^x+5=0
  
  = (e^x)^2-6*e^x+5=0
  
  = (e^x)^2-5*e^x - e^x +5=0
  
  = e^x*(e^x - 5) - 1(e^x - 5)=0
  
  = (e^x - 1)*(e^x - 5)=0
  
  e^x - 1 = 0
  
  e^x = 1
  
  This is the case when x = 0
  
  e^x - 5 = 0
  
  e^x = 5
  
  Take the natural logarithm of both the sides
  
  x = ln 5
  
  The solution of the equation e^2x-6*e^x+5=0 is x = 0 and x = ln 5
  
  giorgiana1976 | College Teacher | (Level 3) Valedictorian
  
  Posted on
  
  We'll solve the equation using the technique of substitution.
  
  We'll substitute e^x by the variable t.
  
  e^2x=(e^x)^2=t^2
  
  We'll re-write the equation in t:
  
  t^2 -6t + 5 = 0
  
  t1 = [6+sqrt(36-20)]/2
  
  t1 = (6+4)/2
  
  t1 = 5
  
  t2 = (6-4)/2
  
  t2 = 1
  
  Now, we'll determine x:
  
  e^x = t1
  
  e^x = 5
  
  We'll apply natural logarithms boths sides:
  
  ln e^x = ln 5
  
  x = ln 5
  
  e^x = t2
  
  e^x = 1
  
  But e^0 = 1
  
  e^x = e^0
  
  x = 0
  
  The requested solutions of the equation are: {0 ; ln 5}.