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March 1st, 2014, 03:29 AM  #1 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  problem with simplification
3^2/(2.3^n) = 1/{2.3^(n2)} how did we do that? 
March 1st, 2014, 03:58 AM  #2 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: problem with simplification
Take an example where n = 6: In general, the 3^2 on the top line cancels out 3^2 on the bottom, reducing the exponent on the bottom line from n to n2: 
March 1st, 2014, 04:14 AM  #3 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  Re: problem with simplification
I guess I am missing something here; where did 3^2 at the bottom came from or when we introduce it what law are we using?

March 1st, 2014, 04:36 AM  #4 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: problem with simplification Do you know how to do this calculation? 
March 1st, 2014, 05:43 AM  #5 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  Re: problem with simplification
yes. same bases u add the exponents

March 1st, 2014, 06:01 AM  #6 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: problem with simplification
So: 
March 1st, 2014, 06:04 AM  #7 
Member Joined: Sep 2013 Posts: 75 Thanks: 0  Re: problem with simplification
If we were to solve 9^x/ 3^(x1) I wud say 3^{2x (x1)} tel me if this is wrong? Now 3^2/2.3^n if I say 3^(2  n) am wrong how? My question is when we divide and we have same bases we subtract top from bottom or bottom from top To clarify this if we have 3^2÷ 3^n which one is correct 3^(2n) or 3^(n2) Jst confused bear with me 
March 1st, 2014, 06:17 AM  #8 
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: problem with simplification 
March 1st, 2014, 06:58 AM  #9 
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Re: problem with simplification Hello, bongantedd! Mystery solved! Those are not decimal points . . . they are multiplications! 

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