Finding the intersections between $y = e^x$ and $y = x + 2$ algebraically?












5












$begingroup$


In trying to find the intersections between $y = e^x$ and $y = x + 2$ in terms of $x$, I came up with the equation,



$e^x = x + 2$



and subsequently,



$x = ln(x+2)$.



Beyond that point, I am stumped. I am able to solve the equation numerically using a calculator, Newton's method, etc., but need to solve it algebraically. I have done a good deal of research on how to solve this type of problem, but have been unable to find any problems similar enough to be of help.



Thanks to the StackExchange community for your help. I love your sites and have been happy to find answers to hundreds of my own questions on them.






logarithms exponential-function transcendental-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is no closed form unless we use special functions.
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:25










  • $begingroup$
    Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:28










  • $begingroup$
    Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:31










  • $begingroup$
    I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:34










  • $begingroup$
    You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
    $endgroup$
    – NoChance
    Mar 8 '16 at 2:49


















5












$begingroup$


In trying to find the intersections between $y = e^x$ and $y = x + 2$ in terms of $x$, I came up with the equation,



$e^x = x + 2$



and subsequently,



$x = ln(x+2)$.



Beyond that point, I am stumped. I am able to solve the equation numerically using a calculator, Newton's method, etc., but need to solve it algebraically. I have done a good deal of research on how to solve this type of problem, but have been unable to find any problems similar enough to be of help.



Thanks to the StackExchange community for your help. I love your sites and have been happy to find answers to hundreds of my own questions on them.






logarithms exponential-function transcendental-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    There is no closed form unless we use special functions.
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:25










  • $begingroup$
    Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:28










  • $begingroup$
    Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:31










  • $begingroup$
    I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:34










  • $begingroup$
    You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
    $endgroup$
    – NoChance
    Mar 8 '16 at 2:49
















5












5








5





$begingroup$


In trying to find the intersections between $y = e^x$ and $y = x + 2$ in terms of $x$, I came up with the equation,



$e^x = x + 2$



and subsequently,



$x = ln(x+2)$.



Beyond that point, I am stumped. I am able to solve the equation numerically using a calculator, Newton's method, etc., but need to solve it algebraically. I have done a good deal of research on how to solve this type of problem, but have been unable to find any problems similar enough to be of help.



Thanks to the StackExchange community for your help. I love your sites and have been happy to find answers to hundreds of my own questions on them.






logarithms exponential-function transcendental-equations







share|cite|improve this question











$endgroup$




In trying to find the intersections between $y = e^x$ and $y = x + 2$ in terms of $x$, I came up with the equation,



$e^x = x + 2$



and subsequently,



$x = ln(x+2)$.



Beyond that point, I am stumped. I am able to solve the equation numerically using a calculator, Newton's method, etc., but need to solve it algebraically. I have done a good deal of research on how to solve this type of problem, but have been unable to find any problems similar enough to be of help.



Thanks to the StackExchange community for your help. I love your sites and have been happy to find answers to hundreds of my own questions on them.






logarithms exponential-function transcendental-equations




logarithms exponential-function transcendental-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 10:24









Martin Sleziak

44.8k10119272




44.8k10119272










asked Mar 8 '16 at 2:23









Dillon MorseDillon Morse

283




283












  • $begingroup$
    There is no closed form unless we use special functions.
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:25










  • $begingroup$
    Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:28










  • $begingroup$
    Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:31










  • $begingroup$
    I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:34










  • $begingroup$
    You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
    $endgroup$
    – NoChance
    Mar 8 '16 at 2:49




















  • $begingroup$
    There is no closed form unless we use special functions.
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:25










  • $begingroup$
    Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:28










  • $begingroup$
    Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
    $endgroup$
    – MathematicsStudent1122
    Mar 8 '16 at 2:31










  • $begingroup$
    I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 2:34










  • $begingroup$
    You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
    $endgroup$
    – NoChance
    Mar 8 '16 at 2:49


















$begingroup$
There is no closed form unless we use special functions.
$endgroup$
– MathematicsStudent1122
Mar 8 '16 at 2:25




$begingroup$
There is no closed form unless we use special functions.
$endgroup$
– MathematicsStudent1122
Mar 8 '16 at 2:25












$begingroup$
Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
$endgroup$
– Dillon Morse
Mar 8 '16 at 2:28




$begingroup$
Could you elaborate a bit? I recall running into Lambert functions earlier. Do you mean that I must use functions such as those to solve for $x$, rather than simple algebraic rules?
$endgroup$
– Dillon Morse
Mar 8 '16 at 2:28












$begingroup$
Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
$endgroup$
– MathematicsStudent1122
Mar 8 '16 at 2:31




$begingroup$
Yes. See: wolframalpha.com/input/?i=e%5Ex+-+x+-+2+%3D+0
$endgroup$
– MathematicsStudent1122
Mar 8 '16 at 2:31












$begingroup$
I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
$endgroup$
– Dillon Morse
Mar 8 '16 at 2:34




$begingroup$
I see. Thank you for the help. I was hoping to be able to solve this algebraically, but it seems my calculus professor may have made a mistake by labeling this as a no-calculator problem.
$endgroup$
– Dillon Morse
Mar 8 '16 at 2:34












$begingroup$
You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
$endgroup$
– NoChance
Mar 8 '16 at 2:49






$begingroup$
You could try to expand $e^x$ using Taylor Series for some terms. However, this will give you an approximate solution. A similar (unanswered so far question) is here:math.stackexchange.com/questions/1686045/…
$endgroup$
– NoChance
Mar 8 '16 at 2:49












1 Answer
1






active

oldest

votes


















3












$begingroup$

Here again appears the beautiful Lambert function : rewrite $$e^x=x+2implies e^{x+2}=e^2(x+2)implies e^y=e^2 y$$ and the solutions are given by $$x_1=-Wleft(-frac{1}{e^2}right)-2$$ $$x_2=-W_{-1}left(-frac{1}{e^2}right)-2$$ In fact, keep in mind that any equation which can write $A+Bx+Clog(D+Ex)=0$ has solutions in terms of Lambert function.



The Wikipedia page gives series approximations.



There is no other closed form to this equation. If you cannot use it, just numerical methods will provide the solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 11:36











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%2f1687866%2ffinding-the-intersections-between-y-ex-and-y-x-2-algebraically%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









3












$begingroup$

Here again appears the beautiful Lambert function : rewrite $$e^x=x+2implies e^{x+2}=e^2(x+2)implies e^y=e^2 y$$ and the solutions are given by $$x_1=-Wleft(-frac{1}{e^2}right)-2$$ $$x_2=-W_{-1}left(-frac{1}{e^2}right)-2$$ In fact, keep in mind that any equation which can write $A+Bx+Clog(D+Ex)=0$ has solutions in terms of Lambert function.



The Wikipedia page gives series approximations.



There is no other closed form to this equation. If you cannot use it, just numerical methods will provide the solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 11:36
















3












$begingroup$

Here again appears the beautiful Lambert function : rewrite $$e^x=x+2implies e^{x+2}=e^2(x+2)implies e^y=e^2 y$$ and the solutions are given by $$x_1=-Wleft(-frac{1}{e^2}right)-2$$ $$x_2=-W_{-1}left(-frac{1}{e^2}right)-2$$ In fact, keep in mind that any equation which can write $A+Bx+Clog(D+Ex)=0$ has solutions in terms of Lambert function.



The Wikipedia page gives series approximations.



There is no other closed form to this equation. If you cannot use it, just numerical methods will provide the solutions.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 11:36














3












3








3





$begingroup$

Here again appears the beautiful Lambert function : rewrite $$e^x=x+2implies e^{x+2}=e^2(x+2)implies e^y=e^2 y$$ and the solutions are given by $$x_1=-Wleft(-frac{1}{e^2}right)-2$$ $$x_2=-W_{-1}left(-frac{1}{e^2}right)-2$$ In fact, keep in mind that any equation which can write $A+Bx+Clog(D+Ex)=0$ has solutions in terms of Lambert function.



The Wikipedia page gives series approximations.



There is no other closed form to this equation. If you cannot use it, just numerical methods will provide the solutions.






share|cite|improve this answer











$endgroup$



Here again appears the beautiful Lambert function : rewrite $$e^x=x+2implies e^{x+2}=e^2(x+2)implies e^y=e^2 y$$ and the solutions are given by $$x_1=-Wleft(-frac{1}{e^2}right)-2$$ $$x_2=-W_{-1}left(-frac{1}{e^2}right)-2$$ In fact, keep in mind that any equation which can write $A+Bx+Clog(D+Ex)=0$ has solutions in terms of Lambert function.



The Wikipedia page gives series approximations.



There is no other closed form to this equation. If you cannot use it, just numerical methods will provide the solutions.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 8 '16 at 6:49

























answered Mar 8 '16 at 6:41









Claude LeiboviciClaude Leibovici

123k1157134




123k1157134












  • $begingroup$
    Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 11:36


















  • $begingroup$
    Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
    $endgroup$
    – Dillon Morse
    Mar 8 '16 at 11:36
















$begingroup$
Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
$endgroup$
– Dillon Morse
Mar 8 '16 at 11:36




$begingroup$
Thanks for the answer. I've marked your answer as accepted. While it does not technically answer my question, it does provide the best possible answer to it.
$endgroup$
– Dillon Morse
Mar 8 '16 at 11:36


















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%2f1687866%2ffinding-the-intersections-between-y-ex-and-y-x-2-algebraically%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