My Math Forum simple, yet insane!....for me at least

 Algebra Pre-Algebra and Basic Algebra Math Forum

December 7th, 2009, 02:39 PM   #1
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Joined: Nov 2009

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simple, yet insane!....for me at least

ok guys,

i know the answer just gotta show me the method in layman terms

THANKS,
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 December 7th, 2009, 04:26 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: simple, yet insane!....for me at least Since you're going to be asked at some point, I might as well do it... What have you done so far? Can you write each factor in #5 as a power of 2? How can you get the factors in #6 that aren't 6 to be 6?
 December 7th, 2009, 04:32 PM #3 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: simple, yet insane!....for me at least Hi; for #5 the answer is 2^(2n-4) $2^n \ * \ 8^n= 2^n \ * \ 2^{3n} = 2^{4n}$ $2^{2n} \ *\ 16= 2^{2n} \ * \ 2^4 = 2^{2n+4}$ $\frac{2^{4n}}{2^{2n+4}}=2^{2n-4}$ for # 6 the answer is 6^(3x)
 December 7th, 2009, 06:04 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,753 Thanks: 2136 $6.\text{ }2^{-x}\,\times\,3^{-x}\,\times\,6^{2x}\,\times\,3^{2x}\,\times\,2^{2x} \,=\,(2\,\times\,3)^{-x}\,\times\,6^{2x}\,\times\,(3\,\times\,2)^{2x}\,= \,6^{-x\,+2x\,+\,2x}\,=\,6^{3x}.$
 December 8th, 2009, 12:32 AM #5 Member   Joined: Nov 2009 Posts: 49 Thanks: 0 Re: simple, yet insane!....for me at least thanks for all the help guys
 December 8th, 2009, 08:40 AM #6 Senior Member   Joined: Dec 2009 From: Las Vegas Posts: 209 Thanks: 0 Re: simple, yet insane!....for me at least Hi peterle1; Glad to help! That you understood the methods is even better. Working with exponents is really a lot of fun.

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