text: # Verify if exponential equation has any roots. 4^x=20-4^(3-x)
  
  hala718 | Certified Educator
  
  4^x = 20 - 4^(3-x)
  
  We need to find the values of x, if any, that satisfies the exponential equation.
  
  First we will rewrite 4^(3-x) = 4^3/4^x
  
  ==> 4^x = 20 - 4^3/4^x
  
  Now we will assume that :
  
  y= 4^x
  
  ==> y = 20 -  64/y
  
  Now we will multiply by y.
  
  ==> y^2 = 20y - 64
  
  ==> y^2 - 20 + 64 = 0
  
  Now we will factor.
  
  ==> (y-16)(y-4) = 0
  
  ==> y1= 16 ==> 4^x = 16 = 4^2 ==> x1 = 2
  
  ==> y2= 4 ==> 4^x = 4 = 4^1 ==> x2= 1
  
  Then we have two values for x that satisfies the equation:
  
  x = { 1, 2}
  
  giorgiana1976 | Student
  
  We'll re-write the term 4^(3-x) recalling the quotient property of exponentials:
  
  4^(3-x) = 4^3/4^x
  
  We'll re-write the equation:
  
  4^3/4^x + 4^x - 20 = 0
  
  We'll multiply by 4^x both sides:
  
  4^3 + 4^2x - 20*4^x = 0
  
  We'll substitute 4^x = t:
  
  t^2 - 20t + 64 = 0
  
  t1 = [20+sqrt(400 - 256)]/2
  
  t1 = (20+sqrt144)/2
  
  t1 = (20+12)/2
  
  t1 = 16
  
  t2 = (20-12)/2
  
  t2 = 4
  
  But t1 = 4^x => 16 = 4^x
  
  We'll put 16 = 4^2
  
  4^x = 4^2
  
  Since the bases are matching, we'll apply one to one property:
  
  x = 2
  
  t2 = 4 => 4^x = 4 => x = 1
  
  The equation has 2 real solutions and they are: {1 ; 2}.
explanation: The document contains a math question and its solution. The question is about finding the roots of the exponential equation \(4^x = 20 - 4^{3-x}\). The document provides a detailed step-by-step solution, leading to the conclusion that the solutions are \(x = 1\) and \(x = 2\).
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distilabel_metadata.content: # Verify if exponential equation has any roots. 4^x=20-4^(3-x)
  
  hala718 | Certified Educator
  
  4^x = 20 - 4^(3-x)
  
  We need to find the values of x, if any, that satisfies the exponential equation.
  
  First we will rewrite 4^(3-x) = 4^3/4^x
  
  ==> 4^x = 20 - 4^3/4^x
  
  Now we will assume that :
  
  y= 4^x
  
  ==> y = 20 -  64/y
  
  Now we will multiply by y.
  
  ==> y^2 = 20y - 64
  
  ==> y^2 - 20 + 64 = 0
  
  Now we will factor.
  
  ==> (y-16)(y-4) = 0
  
  ==> y1= 16 ==> 4^x = 16 = 4^2 ==> x1 = 2
  
  ==> y2= 4 ==> 4^x = 4 = 4^1 ==> x2= 1
  
  Then we have two values for x that satisfies the equation:
  
  x = { 1, 2}
  
  giorgiana1976 | Student
  
  We'll re-write the term 4^(3-x) recalling the quotient property of exponentials:
  
  4^(3-x) = 4^3/4^x
  
  We'll re-write the equation:
  
  4^3/4^x + 4^x - 20 = 0
  
  We'll multiply by 4^x both sides:
  
  4^3 + 4^2x - 20*4^x = 0
  
  We'll substitute 4^x = t:
  
  t^2 - 20t + 64 = 0
  
  t1 = [20+sqrt(400 - 256)]/2
  
  t1 = (20+sqrt144)/2
  
  t1 = (20+12)/2
  
  t1 = 16
  
  t2 = (20-12)/2
  
  t2 = 4
  
  But t1 = 4^x => 16 = 4^x
  
  We'll put 16 = 4^2
  
  4^x = 4^2
  
  Since the bases are matching, we'll apply one to one property:
  
  x = 2
  
  t2 = 4 => 4^x = 4 => x = 1
  
  The equation has 2 real solutions and they are: {1 ; 2}.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its solution. The question is about finding the roots of the exponential equation \(4^x = 20 - 4^{3-x}\). The document provides a detailed step-by-step solution, leading to the conclusion that the solutions are \(x = 1\) and \(x = 2\).
  
  Label: question-and-answer