Prove that $3^1004+2^2009-3^502cdot 2^1005gt 2009^182$

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Prove that $3^1004+2^2009-3^502cdot 2^1005gt 2009^182$.




My try:



we have $$2^11gt 2009$$



Taking power of $182$ both sides we get



$$2^2002 gt 2009^182$$



Now



$$left(3^1004right)+left(2^2009right)-left(3^502right)left(2^1005right)=2^2002+(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)=2^2002+A$$



Now it suffices to prove $$A gt 0$$ where



$$A=(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)$$



any hint?




algebra-precalculus inequality




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  • I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
    – mrtaurho
    Jul 22 at 17:55










  • youtube.com/watch?v=8-9scNP5KWk
    – Ed Pegg
    Jul 22 at 17:55














up vote
2
down vote

favorite
2













Prove that $3^1004+2^2009-3^502cdot 2^1005gt 2009^182$.




My try:



we have $$2^11gt 2009$$



Taking power of $182$ both sides we get



$$2^2002 gt 2009^182$$



Now



$$left(3^1004right)+left(2^2009right)-left(3^502right)left(2^1005right)=2^2002+(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)=2^2002+A$$



Now it suffices to prove $$A gt 0$$ where



$$A=(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)$$



any hint?




algebra-precalculus inequality




share|cite|improve this question





















  • I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
    – mrtaurho
    Jul 22 at 17:55










  • youtube.com/watch?v=8-9scNP5KWk
    – Ed Pegg
    Jul 22 at 17:55












up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






Prove that $3^1004+2^2009-3^502cdot 2^1005gt 2009^182$.




My try:



we have $$2^11gt 2009$$



Taking power of $182$ both sides we get



$$2^2002 gt 2009^182$$



Now



$$left(3^1004right)+left(2^2009right)-left(3^502right)left(2^1005right)=2^2002+(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)=2^2002+A$$



Now it suffices to prove $$A gt 0$$ where



$$A=(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)$$



any hint?




algebra-precalculus inequality




share|cite|improve this question














Prove that $3^1004+2^2009-3^502cdot 2^1005gt 2009^182$.




My try:



we have $$2^11gt 2009$$



Taking power of $182$ both sides we get



$$2^2002 gt 2009^182$$



Now



$$left(3^1004right)+left(2^2009right)-left(3^502right)left(2^1005right)=2^2002+(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)=2^2002+A$$



Now it suffices to prove $$A gt 0$$ where



$$A=(127)2^2002+left(3^1004right)-left(3^502right)left(2^1005right)$$



any hint?





algebra-precalculus inequality





share|cite|improve this question












share|cite|improve this question




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edited Jul 22 at 17:53









Cave Johnson

3,5871326




3,5871326









asked Jul 22 at 17:50









Ekaveera Kumar Sharma

5,15611122




5,15611122











  • I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
    – mrtaurho
    Jul 22 at 17:55










  • youtube.com/watch?v=8-9scNP5KWk
    – Ed Pegg
    Jul 22 at 17:55
















  • I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
    – mrtaurho
    Jul 22 at 17:55










  • youtube.com/watch?v=8-9scNP5KWk
    – Ed Pegg
    Jul 22 at 17:55















I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
– mrtaurho
Jul 22 at 17:55




I am not sure if this helps but the L.H.S. almost equals a binom. To be exact $(3^502-2^1005)^2=3^1004+2^2010-2cdot 3^5022^1005$
– mrtaurho
Jul 22 at 17:55












youtube.com/watch?v=8-9scNP5KWk
– Ed Pegg
Jul 22 at 17:55




youtube.com/watch?v=8-9scNP5KWk
– Ed Pegg
Jul 22 at 17:55










1 Answer
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up vote
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accepted










$beginalign
3^1004+2^2009-3^502cdot 2^1005 &= 3^1004+2^2008-3^502cdot 2^1005 +2^2008\
&= (2^1004-3^502)^2 +2^2008\
&gt 2^2008\
&gt 2009^182 & smalltext(per the OP argument)\
endalign$






share|cite|improve this answer























  • Crystal clear solution, thanks
    – Ekaveera Kumar Sharma
    Jul 22 at 18:12






  • 1




    Why is it intuitively obvious that $2^2008 > 2009^182$?
    – fleablood
    Jul 22 at 18:16






  • 1




    @fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
    – MathOverview
    Jul 22 at 18:21







  • 1




    Also the OP's post already mentions that $2^2002 > 2009^182$
    – Sil
    Jul 22 at 18:26










  • Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
    – fleablood
    Jul 22 at 18:28










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1 Answer
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active

oldest

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up vote
8
down vote



accepted










$beginalign
3^1004+2^2009-3^502cdot 2^1005 &= 3^1004+2^2008-3^502cdot 2^1005 +2^2008\
&= (2^1004-3^502)^2 +2^2008\
&gt 2^2008\
&gt 2009^182 & smalltext(per the OP argument)\
endalign$






share|cite|improve this answer























  • Crystal clear solution, thanks
    – Ekaveera Kumar Sharma
    Jul 22 at 18:12






  • 1




    Why is it intuitively obvious that $2^2008 > 2009^182$?
    – fleablood
    Jul 22 at 18:16






  • 1




    @fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
    – MathOverview
    Jul 22 at 18:21







  • 1




    Also the OP's post already mentions that $2^2002 > 2009^182$
    – Sil
    Jul 22 at 18:26










  • Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
    – fleablood
    Jul 22 at 18:28














up vote
8
down vote



accepted










$beginalign
3^1004+2^2009-3^502cdot 2^1005 &= 3^1004+2^2008-3^502cdot 2^1005 +2^2008\
&= (2^1004-3^502)^2 +2^2008\
&gt 2^2008\
&gt 2009^182 & smalltext(per the OP argument)\
endalign$






share|cite|improve this answer























  • Crystal clear solution, thanks
    – Ekaveera Kumar Sharma
    Jul 22 at 18:12






  • 1




    Why is it intuitively obvious that $2^2008 > 2009^182$?
    – fleablood
    Jul 22 at 18:16






  • 1




    @fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
    – MathOverview
    Jul 22 at 18:21







  • 1




    Also the OP's post already mentions that $2^2002 > 2009^182$
    – Sil
    Jul 22 at 18:26










  • Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
    – fleablood
    Jul 22 at 18:28












up vote
8
down vote



accepted







up vote
8
down vote



accepted






$beginalign
3^1004+2^2009-3^502cdot 2^1005 &= 3^1004+2^2008-3^502cdot 2^1005 +2^2008\
&= (2^1004-3^502)^2 +2^2008\
&gt 2^2008\
&gt 2009^182 & smalltext(per the OP argument)\
endalign$






share|cite|improve this answer















$beginalign
3^1004+2^2009-3^502cdot 2^1005 &= 3^1004+2^2008-3^502cdot 2^1005 +2^2008\
&= (2^1004-3^502)^2 +2^2008\
&gt 2^2008\
&gt 2009^182 & smalltext(per the OP argument)\
endalign$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 22 at 20:11


























answered Jul 22 at 18:10









Joffan

31.8k43169




31.8k43169











  • Crystal clear solution, thanks
    – Ekaveera Kumar Sharma
    Jul 22 at 18:12






  • 1




    Why is it intuitively obvious that $2^2008 > 2009^182$?
    – fleablood
    Jul 22 at 18:16






  • 1




    @fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
    – MathOverview
    Jul 22 at 18:21







  • 1




    Also the OP's post already mentions that $2^2002 > 2009^182$
    – Sil
    Jul 22 at 18:26










  • Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
    – fleablood
    Jul 22 at 18:28
















  • Crystal clear solution, thanks
    – Ekaveera Kumar Sharma
    Jul 22 at 18:12






  • 1




    Why is it intuitively obvious that $2^2008 > 2009^182$?
    – fleablood
    Jul 22 at 18:16






  • 1




    @fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
    – MathOverview
    Jul 22 at 18:21







  • 1




    Also the OP's post already mentions that $2^2002 > 2009^182$
    – Sil
    Jul 22 at 18:26










  • Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
    – fleablood
    Jul 22 at 18:28















Crystal clear solution, thanks
– Ekaveera Kumar Sharma
Jul 22 at 18:12




Crystal clear solution, thanks
– Ekaveera Kumar Sharma
Jul 22 at 18:12




1




1




Why is it intuitively obvious that $2^2008 > 2009^182$?
– fleablood
Jul 22 at 18:16




Why is it intuitively obvious that $2^2008 > 2009^182$?
– fleablood
Jul 22 at 18:16




1




1




@fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
– MathOverview
Jul 22 at 18:21





@fleablood Note that $2^2008=(2^11)^frac200811=(2048)^186.545454...>(2048)^186>(2009)^186$.
– MathOverview
Jul 22 at 18:21





1




1




Also the OP's post already mentions that $2^2002 > 2009^182$
– Sil
Jul 22 at 18:26




Also the OP's post already mentions that $2^2002 > 2009^182$
– Sil
Jul 22 at 18:26












Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
– fleablood
Jul 22 at 18:28




Yeah.... sure.... It just... I dunno.... as a problem it lack elegance. I mean.... what's the point? "Also the OP's post already mentions that 22002>2009182 " Oh, why so it does. Question withdrawn.
– fleablood
Jul 22 at 18:28












 

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