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

45.2k11123278




45.2k11123278










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












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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

127k1158135




127k1158135












  • $begingroup$
    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


















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