text: PS-Exponents : PS Archive
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  # PS-Exponents
  
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  20 Oct 2005, 00:55
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  My friend gave me to solve. Don't know the answer choices.
  
  3^6x=8100
  
  (3^(x-1))^3=?
  
  [edited by Hong to correct the question stem as noted below.]
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  20 Oct 2005, 02:17
  i dont see the trick i can't find the path
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  20 Oct 2005, 02:50
  I don't know the official answer but I still gave it a try. Just don't know how far it is correct.
  
  Since 3^6x=8100
  
  3^6 * 3^x=8100
  
  3^x=8100/3^6
  =8100/243
  =100/3
  
  Substituting value of 3^x in the equation (3^x-1)^3
  
  ((100/3)/(1/3))^3
  
  (100/9)^3.
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  20 Oct 2005, 02:55
  getzgetzu wrote:
  My friend gave me to solve. Don't know the answer choices.
  
  3^6x=8100
  
  (3^x-1)^3=?
  
  use the formula (a-b)^3= a^3 - 3a^2*b + 3 a*b^2 - b^3
  
  (3^x-1)^3= 3^(3x) - 3*3^(2x) + 3*3^x -1
  =sqrt(3^(6x))- 3* cubic root of 3^(6x) + 3* 6th root of 3^(6x)-1
  = 90+ 60.25 ( rounded) + 13.45 (rounded)-1
  = 162.7
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  20 Oct 2005, 02:57
  Can somebody help me where did I wrong? Please.
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  20 Oct 2005, 02:58
  see my explanation above ...hope this helps
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  20 Oct 2005, 03:09
  I am extremely sorry for the typo
  
  3^6x=8100
  
  (3^(x-1))^3=?
  
  So we cannot use (a-b)^3 formula here. I wish we had MathTypeâ„¢, we would have avoided all such errors.
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  20 Oct 2005, 03:16
  getzgetzu wrote:
  I am extremely sorry for the typo
  
  3^6x=8100
  
  (3^(x-1))^3=?
  
  So we cannot use (a-b)^3 formula here. I wish we had MathTypeâ„¢, we would have avoided all such errors.
  
  It's even easier!
  
  (3^(x-1))^3= 3^(3*(x-1)) = 3^(3x-3)= (3^3x)/ 3^3= sqrt(3^(6x))/27=90/27= 10/3
  20 Oct 2005, 03:16
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