Ambiguous solutions to the same exponential problem - help needed












1












$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















1












$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














1












1








1





$begingroup$


I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality







share|cite|improve this question











$endgroup$




I came across the below question in an IMO preparatory book and tried to solve it, I followed a certain steps and got a solution, but the book followed a slightly different approach and arrived at a totally different solution, I couldn't find any fault with both approaches.



The problems is as follows:




Find the least value of $a$ for which $5^{1+x}+5^{1-x}$, $a/2$, $25^x+25^{-x}$ are three consecutive terms of an Arithmetic Progression.




My approach :



since the three terms are to be in AP



$a/2 = (5^{1+x}+5^{1-x}+25^x+25^{-x})/2$



$Rightarrow a = 5cdot5^x+5cdot5^{-x}+5^{2x}+5^{-2x}$. Let $5^x = t$ (Step A)



Then $a = 5t+(5/t)+t^2+t^{-2}$. Now let $u=t+1/t$



$Rightarrow a = 5u+u^2-2$, since $u^2 = t^2+t^{-2}+2$ (Step B)



This is a quadratic $u^2+5u-(2+a) = 0$



for $u$ to be real $25+4(2+a) ge 0$,



$Rightarrow a ge -33/4$.



I even tried maximizing the expression on the RHS in Step B , i.e. maximizing $u^2+5u-2$, then differentiating and equating to zero $2u+5 = 0$,



$Rightarrow u = -5/2$, substituting this value in the expression



$(-5/2)^2+5(-5/2)-2 = -33/4$ Same solution as above



Approach followed in the book below.



It's the same up to step A mentioned above.



then the equation becomes



$a = (t-(1/t))^2 +5(sqrt{t}-1/sqrt{t}))^2 +12$



since there are two square expressions, the value of this expression is always $ge12$, hence $age12$.



Both approaches seem fine to me, can someone explain the ambiguity?






inequality




inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 9:53









egreg

183k1486205




183k1486205










asked Jan 6 at 9:34









Madavan ViswanathanMadavan Viswanathan

365




365








  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














  • 1




    $begingroup$
    In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
    $endgroup$
    – Mindlack
    Jan 6 at 9:44










  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11








1




1




$begingroup$
In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
$endgroup$
– Mindlack
Jan 6 at 9:44




$begingroup$
In your approach, $t$ can be any positive real number, thus $u$ has to be in $[2,infty]$. So he condition is not that $Delta geq 0$, but: there is a solution that is $geq 2$.
$endgroup$
– Mindlack
Jan 6 at 9:44












$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11




$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11










1 Answer
1






active

oldest

votes


















1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11











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',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
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%2f3063647%2fambiguous-solutions-to-the-same-exponential-problem-help-needed%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















1












$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11














1












1








1





$begingroup$

The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$






share|cite|improve this answer









$endgroup$



The expression $u=t+t^{-1}$ cannot take any value. First of all, $t=5^x>0$; the minimum of $u=t+1/t$ is taken at $t=1$ (AM-GM); hence $t+1/tge2$. All values $ge2$ for $u$ can be obtained.



You then have
$$
a=u^2+5u-2ge 4+10-2=12
$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 6 at 9:58









egregegreg

183k1486205




183k1486205












  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11


















  • $begingroup$
    ok ,got it , Thanks very much.
    $endgroup$
    – Madavan Viswanathan
    Jan 6 at 10:11
















$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11




$begingroup$
ok ,got it , Thanks very much.
$endgroup$
– Madavan Viswanathan
Jan 6 at 10:11


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3063647%2fambiguous-solutions-to-the-same-exponential-problem-help-needed%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
Read more

build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

Image
0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
Read more