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

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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













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




share|cite|improve this question








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
1






active

oldest

votes

















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










Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859624%2fprove-that-3100422009-3502-cdot-21005-gt-2009182%23new-answer', 'question_page');

);

Post as a guest

















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








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












 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2859624%2fprove-that-3100422009-3502-cdot-21005-gt-2009182%23new-answer', 'question_page');

);

Post as a guest
































































Comments

Post a Comment

Popular posts from this blog

What is the equation of a 3D cone with generalised tilt?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite What is the equation of a 3D cone with generalized tilt? I've noticed that in most equations given to represent a cone, there is no parameter which defines the tilt of the cone in 3D space and that most of them have their point at the origin $(0,0,0)$ - I was wondering if anyone could give me a more generalized cone equation for a cone in any position in 3D space 3d share | cite | improve this question edited Jul 25 '16 at 0:05 ervx 9,425 3 13 36 asked Jul 24 '16 at 15:38 Charlene 11 2 2 Welcome to Math.SE! Could you please explain a few things to help people give an answer useful to you: 1. What do you mean by "generalized tilt"? 2. Are you asking about a right circular cone? Does the angle at the vertex matter? 3. What form of answer (implicit, parametric...?) are you looking for? 4. If this is homework, what tools do you have available, an
Read more

Color the edges and diagonals of a regular polygon

Image
Clash Royale CLAN TAG #URR8PPP up vote 5 down vote favorite 1 Here is the problem: For what $n$ is it possible to color the edges and diagonals of an $n$-side regular polygon with $dfracbinomn23$ colors, such that you use every color exactly three times and for every color the three segments (edges or diagonals) with that color form a triangle? Trivially $nequiv 0 text or 1 pmod3$. I can also prove that if the statement is true for $k$ than it is true for $3k$ as well. How to finish? Please help! Thanks combinatorial-geometry share | cite | improve this question edited Aug 6 at 21:57 alcana 118 4 asked Aug 6 at 21:15 Leo Gardner 356 11 Even assuming "polyhpn" in the title is a typo for polygon , it is unclear what you mean by "the edges and sides". Aren't the side of a regular polygon the same as the edges? A regular $n$-sided polygon has $n$ edges. – hardmath Aug 6 at 21:20 3 $n$ ne
Read more

Relationship between determinant of matrix and determinant of adjoint?

Image
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite We are studying adjoints in class, and I was curious if there is a relationship between the determinant of matrix A, and the determinant of the adjoint of matrix A? I assume there would be a relationship because finding the adjoint requires creating a cofactor matrix and then transposing it. linear-algebra determinant adjoint-operators share | cite | improve this question asked Nov 17 '17 at 17:32 CluelessCoder 32 5 For a square matrix $A$ of order $n$, we have $A(operatornameadj A)=(operatornameadj A)A=|A|I_n$ where $I_n$ is the identity matrix of order $n$. This should tell you about the determinant of the adjoint in terms of that of $A$ (use multiplicative property and other elementary properties of determinants). – Prasun Biswas Nov 17 '17 at 17:35 1 @CluelessCoder: see here: math.stackexchange.com/questions/516127/… – Hector Blandin Nov 17 &
Read more