How can I find y?












2












$begingroup$


The following equation is given: $$ye^y=e^{x+1}$$ when $x=0$. I tried to solve it as logarithmic equation but I can't go further. I know $y=1$ but I don't know how to prove it. Any idea? Thank you😊






exponential-function functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – Matti P.
    12 hours ago






  • 1




    $begingroup$
    You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
    $endgroup$
    – James
    12 hours ago






  • 3




    $begingroup$
    "I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
    $endgroup$
    – Yves Daoust
    12 hours ago






  • 3




    $begingroup$
    There is no real need to introduce $x$ then set it to $0$.
    $endgroup$
    – Yves Daoust
    12 hours ago










  • $begingroup$
    I found it! It's right there at the beginning of your equation!
    $endgroup$
    – Mason Wheeler
    7 hours ago
















2












$begingroup$


The following equation is given: $$ye^y=e^{x+1}$$ when $x=0$. I tried to solve it as logarithmic equation but I can't go further. I know $y=1$ but I don't know how to prove it. Any idea? Thank you😊






exponential-function functional-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – Matti P.
    12 hours ago






  • 1




    $begingroup$
    You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
    $endgroup$
    – James
    12 hours ago






  • 3




    $begingroup$
    "I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
    $endgroup$
    – Yves Daoust
    12 hours ago






  • 3




    $begingroup$
    There is no real need to introduce $x$ then set it to $0$.
    $endgroup$
    – Yves Daoust
    12 hours ago










  • $begingroup$
    I found it! It's right there at the beginning of your equation!
    $endgroup$
    – Mason Wheeler
    7 hours ago














2












2








2


1



$begingroup$


The following equation is given: $$ye^y=e^{x+1}$$ when $x=0$. I tried to solve it as logarithmic equation but I can't go further. I know $y=1$ but I don't know how to prove it. Any idea? Thank you😊






exponential-function functional-equations







share|cite|improve this question











$endgroup$




The following equation is given: $$ye^y=e^{x+1}$$ when $x=0$. I tried to solve it as logarithmic equation but I can't go further. I know $y=1$ but I don't know how to prove it. Any idea? Thank you😊






exponential-function functional-equations




exponential-function functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 12 hours ago









SNEHIL SANYAL

570110




570110










asked 12 hours ago









AnnaAnna

2617




2617












  • $begingroup$
    Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – Matti P.
    12 hours ago






  • 1




    $begingroup$
    You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
    $endgroup$
    – James
    12 hours ago






  • 3




    $begingroup$
    "I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
    $endgroup$
    – Yves Daoust
    12 hours ago






  • 3




    $begingroup$
    There is no real need to introduce $x$ then set it to $0$.
    $endgroup$
    – Yves Daoust
    12 hours ago










  • $begingroup$
    I found it! It's right there at the beginning of your equation!
    $endgroup$
    – Mason Wheeler
    7 hours ago


















  • $begingroup$
    Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – Matti P.
    12 hours ago






  • 1




    $begingroup$
    You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
    $endgroup$
    – James
    12 hours ago






  • 3




    $begingroup$
    "I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
    $endgroup$
    – Yves Daoust
    12 hours ago






  • 3




    $begingroup$
    There is no real need to introduce $x$ then set it to $0$.
    $endgroup$
    – Yves Daoust
    12 hours ago










  • $begingroup$
    I found it! It's right there at the beginning of your equation!
    $endgroup$
    – Mason Wheeler
    7 hours ago
















$begingroup$
Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Matti P.
12 hours ago




$begingroup$
Indeed, someone has thought if this problem before: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– Matti P.
12 hours ago




1




1




$begingroup$
You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
$endgroup$
– James
12 hours ago




$begingroup$
You are saying that $ye^y=e^{x+1}$ holds for $x=0$ so that $ye^y=e^1=e$? But then, $y=1$ obviously is a solution.
$endgroup$
– James
12 hours ago




3




3




$begingroup$
"I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
$endgroup$
– Yves Daoust
12 hours ago




$begingroup$
"I know $y=1$ but I don't know how to prove it": what about $1e^1=e^{0+1}$ ? :-)
$endgroup$
– Yves Daoust
12 hours ago




3




3




$begingroup$
There is no real need to introduce $x$ then set it to $0$.
$endgroup$
– Yves Daoust
12 hours ago




$begingroup$
There is no real need to introduce $x$ then set it to $0$.
$endgroup$
– Yves Daoust
12 hours ago












$begingroup$
I found it! It's right there at the beginning of your equation!
$endgroup$
– Mason Wheeler
7 hours ago




$begingroup$
I found it! It's right there at the beginning of your equation!
$endgroup$
– Mason Wheeler
7 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

You may write




  • $ye^y = e^{ln y + y} = e^{x+1} Rightarrow ln y + y = 1+x stackrel{x= 0}{Rightarrow} Rightarrow ln y + y = 1 Rightarrow y =1$


The uniqueness of the solution follows from the monotonicity of $ln y+ y$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Can't you just say that $ye^y$ is monotonic ?
    $endgroup$
    – Yves Daoust
    6 hours ago










  • $begingroup$
    Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
    $endgroup$
    – trancelocation
    5 hours ago












  • $begingroup$
    I can't understand the need to take the logarithm...
    $endgroup$
    – Yves Daoust
    5 hours ago










  • $begingroup$
    I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
    $endgroup$
    – trancelocation
    5 hours ago










  • $begingroup$
    I mean that I don't see in what way the logarithm simplifies the resolution.
    $endgroup$
    – Yves Daoust
    5 hours ago



















3












$begingroup$

You can solve this equation by taking logarithms both side with respect to the base $e$.
The equation becomes:
$$ln(ye^y)=ln(e^{x+1})$$
$$ln y+ln e^y=ln (e^{x+1})$$
$$ln y +y = x+1$$
You are given the value of $x=0$, put it in the equation to get:
$$ln y +y =1$$



From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.



enter image description here



Hope this helps...






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    "You can solve this equation by taking logarithms both side": this doesn't help.
    $endgroup$
    – Yves Daoust
    12 hours ago



















2












$begingroup$

If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=ecdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.






share|cite|improve this answer









$endgroup$













    Your Answer





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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    You may write




    • $ye^y = e^{ln y + y} = e^{x+1} Rightarrow ln y + y = 1+x stackrel{x= 0}{Rightarrow} Rightarrow ln y + y = 1 Rightarrow y =1$


    The uniqueness of the solution follows from the monotonicity of $ln y+ y$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can't you just say that $ye^y$ is monotonic ?
      $endgroup$
      – Yves Daoust
      6 hours ago










    • $begingroup$
      Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
      $endgroup$
      – trancelocation
      5 hours ago












    • $begingroup$
      I can't understand the need to take the logarithm...
      $endgroup$
      – Yves Daoust
      5 hours ago










    • $begingroup$
      I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
      $endgroup$
      – trancelocation
      5 hours ago










    • $begingroup$
      I mean that I don't see in what way the logarithm simplifies the resolution.
      $endgroup$
      – Yves Daoust
      5 hours ago
















    3












    $begingroup$

    You may write




    • $ye^y = e^{ln y + y} = e^{x+1} Rightarrow ln y + y = 1+x stackrel{x= 0}{Rightarrow} Rightarrow ln y + y = 1 Rightarrow y =1$


    The uniqueness of the solution follows from the monotonicity of $ln y+ y$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Can't you just say that $ye^y$ is monotonic ?
      $endgroup$
      – Yves Daoust
      6 hours ago










    • $begingroup$
      Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
      $endgroup$
      – trancelocation
      5 hours ago












    • $begingroup$
      I can't understand the need to take the logarithm...
      $endgroup$
      – Yves Daoust
      5 hours ago










    • $begingroup$
      I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
      $endgroup$
      – trancelocation
      5 hours ago










    • $begingroup$
      I mean that I don't see in what way the logarithm simplifies the resolution.
      $endgroup$
      – Yves Daoust
      5 hours ago














    3












    3








    3





    $begingroup$

    You may write




    • $ye^y = e^{ln y + y} = e^{x+1} Rightarrow ln y + y = 1+x stackrel{x= 0}{Rightarrow} Rightarrow ln y + y = 1 Rightarrow y =1$


    The uniqueness of the solution follows from the monotonicity of $ln y+ y$.






    share|cite|improve this answer









    $endgroup$



    You may write




    • $ye^y = e^{ln y + y} = e^{x+1} Rightarrow ln y + y = 1+x stackrel{x= 0}{Rightarrow} Rightarrow ln y + y = 1 Rightarrow y =1$


    The uniqueness of the solution follows from the monotonicity of $ln y+ y$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 12 hours ago









    trancelocationtrancelocation

    12.5k1826




    12.5k1826












    • $begingroup$
      Can't you just say that $ye^y$ is monotonic ?
      $endgroup$
      – Yves Daoust
      6 hours ago










    • $begingroup$
      Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
      $endgroup$
      – trancelocation
      5 hours ago












    • $begingroup$
      I can't understand the need to take the logarithm...
      $endgroup$
      – Yves Daoust
      5 hours ago










    • $begingroup$
      I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
      $endgroup$
      – trancelocation
      5 hours ago










    • $begingroup$
      I mean that I don't see in what way the logarithm simplifies the resolution.
      $endgroup$
      – Yves Daoust
      5 hours ago


















    • $begingroup$
      Can't you just say that $ye^y$ is monotonic ?
      $endgroup$
      – Yves Daoust
      6 hours ago










    • $begingroup$
      Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
      $endgroup$
      – trancelocation
      5 hours ago












    • $begingroup$
      I can't understand the need to take the logarithm...
      $endgroup$
      – Yves Daoust
      5 hours ago










    • $begingroup$
      I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
      $endgroup$
      – trancelocation
      5 hours ago










    • $begingroup$
      I mean that I don't see in what way the logarithm simplifies the resolution.
      $endgroup$
      – Yves Daoust
      5 hours ago
















    $begingroup$
    Can't you just say that $ye^y$ is monotonic ?
    $endgroup$
    – Yves Daoust
    6 hours ago




    $begingroup$
    Can't you just say that $ye^y$ is monotonic ?
    $endgroup$
    – Yves Daoust
    6 hours ago












    $begingroup$
    Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
    $endgroup$
    – trancelocation
    5 hours ago






    $begingroup$
    Yes, of course. But I just said that $ln y + y$ is monotonic. :-) Even if this gives me a downvote as I have just seen :-D. Funny.
    $endgroup$
    – trancelocation
    5 hours ago














    $begingroup$
    I can't understand the need to take the logarithm...
    $endgroup$
    – Yves Daoust
    5 hours ago




    $begingroup$
    I can't understand the need to take the logarithm...
    $endgroup$
    – Yves Daoust
    5 hours ago












    $begingroup$
    I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
    $endgroup$
    – trancelocation
    5 hours ago




    $begingroup$
    I thought MSE is a place where the OPs can see different answers to their questions and may decide for themselves what is most helpful for them. The shortest solution is not always the most helpful depending on the knowledge of the OP.
    $endgroup$
    – trancelocation
    5 hours ago












    $begingroup$
    I mean that I don't see in what way the logarithm simplifies the resolution.
    $endgroup$
    – Yves Daoust
    5 hours ago




    $begingroup$
    I mean that I don't see in what way the logarithm simplifies the resolution.
    $endgroup$
    – Yves Daoust
    5 hours ago











    3












    $begingroup$

    You can solve this equation by taking logarithms both side with respect to the base $e$.
    The equation becomes:
    $$ln(ye^y)=ln(e^{x+1})$$
    $$ln y+ln e^y=ln (e^{x+1})$$
    $$ln y +y = x+1$$
    You are given the value of $x=0$, put it in the equation to get:
    $$ln y +y =1$$



    From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.



    enter image description here



    Hope this helps...






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      "You can solve this equation by taking logarithms both side": this doesn't help.
      $endgroup$
      – Yves Daoust
      12 hours ago
















    3












    $begingroup$

    You can solve this equation by taking logarithms both side with respect to the base $e$.
    The equation becomes:
    $$ln(ye^y)=ln(e^{x+1})$$
    $$ln y+ln e^y=ln (e^{x+1})$$
    $$ln y +y = x+1$$
    You are given the value of $x=0$, put it in the equation to get:
    $$ln y +y =1$$



    From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.



    enter image description here



    Hope this helps...






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      "You can solve this equation by taking logarithms both side": this doesn't help.
      $endgroup$
      – Yves Daoust
      12 hours ago














    3












    3








    3





    $begingroup$

    You can solve this equation by taking logarithms both side with respect to the base $e$.
    The equation becomes:
    $$ln(ye^y)=ln(e^{x+1})$$
    $$ln y+ln e^y=ln (e^{x+1})$$
    $$ln y +y = x+1$$
    You are given the value of $x=0$, put it in the equation to get:
    $$ln y +y =1$$



    From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.



    enter image description here



    Hope this helps...






    share|cite|improve this answer









    $endgroup$



    You can solve this equation by taking logarithms both side with respect to the base $e$.
    The equation becomes:
    $$ln(ye^y)=ln(e^{x+1})$$
    $$ln y+ln e^y=ln (e^{x+1})$$
    $$ln y +y = x+1$$
    You are given the value of $x=0$, put it in the equation to get:
    $$ln y +y =1$$



    From the equation, if we put $y=1$ it satisfies the equation. You can also plot a graph for the two functions.



    enter image description here



    Hope this helps...







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 12 hours ago









    SNEHIL SANYALSNEHIL SANYAL

    570110




    570110








    • 2




      $begingroup$
      "You can solve this equation by taking logarithms both side": this doesn't help.
      $endgroup$
      – Yves Daoust
      12 hours ago














    • 2




      $begingroup$
      "You can solve this equation by taking logarithms both side": this doesn't help.
      $endgroup$
      – Yves Daoust
      12 hours ago








    2




    2




    $begingroup$
    "You can solve this equation by taking logarithms both side": this doesn't help.
    $endgroup$
    – Yves Daoust
    12 hours ago




    $begingroup$
    "You can solve this equation by taking logarithms both side": this doesn't help.
    $endgroup$
    – Yves Daoust
    12 hours ago











    2












    $begingroup$

    If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=ecdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=ecdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=ecdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.






        share|cite|improve this answer









        $endgroup$



        If $x=0$, then $y=1$ is obviously a solution. But $e^{x+1}=ecdot e^x>e^x$, and $x+1>x$, hence $xe^x$ is an increasing function. Since $y=1$ solves $ye^y=e$, and $e$ is constant, it is unique.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        YiFanYiFan

        4,3241627




        4,3241627






























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