Find $x$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$












7












$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01


















7












$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01
















7












7








7


0



$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$




Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function




algebra-precalculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 6:57









Dylan

12.9k31027




12.9k31027










asked Dec 26 '18 at 12:54









SandelSandel

1816




1816












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01




















  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01


















$begingroup$
$y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
$endgroup$
– Viktor Glombik
Dec 26 '18 at 13:01






$begingroup$
$y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
$endgroup$
– Viktor Glombik
Dec 26 '18 at 13:01












1 Answer
1






active

oldest

votes


















6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53











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%2f3052915%2ffind-x-such-that-x2-4x-3x-2x-4x-x2-4x-5x%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









6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53
















6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53














6












6








6





$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$



Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 13:06

























answered Dec 26 '18 at 13:01









lab bhattacharjeelab bhattacharjee

225k15157275




225k15157275












  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53


















  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53
















$begingroup$
The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
$endgroup$
– Sandel
Dec 26 '18 at 13:37




$begingroup$
The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
$endgroup$
– Sandel
Dec 26 '18 at 13:37












$begingroup$
@Sandel, For $0le yle1, y^x$ is decreasing right?
$endgroup$
– lab bhattacharjee
Dec 26 '18 at 13:48




$begingroup$
@Sandel, For $0le yle1, y^x$ is decreasing right?
$endgroup$
– lab bhattacharjee
Dec 26 '18 at 13:48












$begingroup$
Yes, I agree with that part.
$endgroup$
– Sandel
Dec 26 '18 at 13:53




$begingroup$
Yes, I agree with that part.
$endgroup$
– Sandel
Dec 26 '18 at 13:53


















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%2f3052915%2ffind-x-such-that-x2-4x-3x-2x-4x-x2-4x-5x%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

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more