text: # Indices (simple but i am have problems)
  
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  • Jan 8th 2012, 09:14 AM
  kingmaths
  Indices (simple but i am getting problems)
  9^x+1 - 13(3^x) + 4 =0
  
  Use y = 3^x
  
  OK so i start and reached to this point
  3^2(x+1) - 13(3^x) + 4 = 0
  
  3^2x+2 - 13(3^x) + 4 = 0
  
  I need some help to proceed, thanks in advance
  • Jan 8th 2012, 09:34 AM
  Bacterius
  Re: Indices (simple but i am have problems)
  $3^{2x + 2} - 13 \times 3^x + 4 = 0$
  
  Expand:
  
  $3^{2x} \times 3^2 - 13 \times 3^x + 4 = 0$
  
  Which becomes (from the rules of indices - look at the first term):
  
  $\left (3^{x} \right )^2 \times 3^2 - 13 \times 3^x + 4 = 0$
  
  Now use the hint and substitute $y= 3^x$
  
  $y^2 \times 3^2 - 13y + 4 = 0$
  
  So:
  
  $9 y^2 - 13y + 4 = 0$
  
  Can you solve for $y$ and put the solution back in terms of $x$ now?
explanation: The document is a math forum post where a user named kingmaths presents a problem involving indices and requests help to proceed with solving the equation \(9^{x+1} - 13(3^x) + 4 = 0\). Another user, Bacterius, provides a detailed step-by-step solution to transform the equation into a quadratic form and suggests solving for \(y\) and then converting the solution back to \(x\).
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Indices (simple but i am have problems)
  
  Printable View
  
  • Jan 8th 2012, 09:14 AM
  kingmaths
  Indices (simple but i am getting problems)
  9^x+1 - 13(3^x) + 4 =0
  
  Use y = 3^x
  
  OK so i start and reached to this point
  3^2(x+1) - 13(3^x) + 4 = 0
  
  3^2x+2 - 13(3^x) + 4 = 0
  
  I need some help to proceed, thanks in advance
  • Jan 8th 2012, 09:34 AM
  Bacterius
  Re: Indices (simple but i am have problems)
  $3^{2x + 2} - 13 \times 3^x + 4 = 0$
  
  Expand:
  
  $3^{2x} \times 3^2 - 13 \times 3^x + 4 = 0$
  
  Which becomes (from the rules of indices - look at the first term):
  
  $\left (3^{x} \right )^2 \times 3^2 - 13 \times 3^x + 4 = 0$
  
  Now use the hint and substitute $y= 3^x$
  
  $y^2 \times 3^2 - 13y + 4 = 0$
  
  So:
  
  $9 y^2 - 13y + 4 = 0$
  
  Can you solve for $y$ and put the solution back in terms of $x$ now?
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a math forum post where a user named kingmaths presents a problem involving indices and requests help to proceed with solving the equation \(9^{x+1} - 13(3^x) + 4 = 0\). Another user, Bacterius, provides a detailed step-by-step solution to transform the equation into a quadratic form and suggests solving for \(y\) and then converting the solution back to \(x\).
  
  Label: question-and-answer