My Math Forum problem with simplification

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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^(n-2)} 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: $\frac{3^2}{2.3^6}= \frac{3^2}{2.3^4.3^2} = \frac{1}{2.3^4}$ 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 n-2: $\frac{3^2}{2.3^n}= \frac{3^2}{2.3^{n-2}.3^2} = \frac{1}{2.3^{n-2}$
 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 $3^2.3^4= ?$ 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: $3^2.3^{n-2}= ?$
 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^(x-1) I wud say 3^{2x -(x-1)} 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^(2-n) or 3^(n-2) 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 $\frac{3^2}{3^n}= \frac{1}{3^{n-2}} = 3^{2-n}$
 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! $\;\;\;\frac{3^2}{2\cdot3^n} \:=\:\frac{1}{2\cdot3^{n-2}}$

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