solving for x in $x^x-x^{-x}=(1+x)^{-x}$












1












$begingroup$


I'm trying to solve $x^x-x^{-x}=(1+x)^{-x}$.
I tried to express $x^x-x^{-x}$ as $2 sinh(x ln x)$ but it didn't seem to make things easier. Any ideas?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $x>0$, $x^x = e^{xln x}$.
    $endgroup$
    – Wuestenfux
    Nov 29 '18 at 15:14






  • 1




    $begingroup$
    $x=-1$ is an obvious solution.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:15












  • $begingroup$
    @AlexSilva Well, watch out for the RHS ...
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 15:24












  • $begingroup$
    $x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
    $endgroup$
    – Lance
    Nov 29 '18 at 15:27












  • $begingroup$
    @MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:30


















1












$begingroup$


I'm trying to solve $x^x-x^{-x}=(1+x)^{-x}$.
I tried to express $x^x-x^{-x}$ as $2 sinh(x ln x)$ but it didn't seem to make things easier. Any ideas?






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    For $x>0$, $x^x = e^{xln x}$.
    $endgroup$
    – Wuestenfux
    Nov 29 '18 at 15:14






  • 1




    $begingroup$
    $x=-1$ is an obvious solution.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:15












  • $begingroup$
    @AlexSilva Well, watch out for the RHS ...
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 15:24












  • $begingroup$
    $x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
    $endgroup$
    – Lance
    Nov 29 '18 at 15:27












  • $begingroup$
    @MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:30
















1












1








1





$begingroup$


I'm trying to solve $x^x-x^{-x}=(1+x)^{-x}$.
I tried to express $x^x-x^{-x}$ as $2 sinh(x ln x)$ but it didn't seem to make things easier. Any ideas?






algebra-precalculus







share|cite|improve this question











$endgroup$




I'm trying to solve $x^x-x^{-x}=(1+x)^{-x}$.
I tried to express $x^x-x^{-x}$ as $2 sinh(x ln x)$ but it didn't seem to make things easier. Any ideas?






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 15:18









axblount

1,99811016




1,99811016










asked Nov 29 '18 at 15:01









Juan Pablo ArcilaJuan Pablo Arcila

412




412












  • $begingroup$
    For $x>0$, $x^x = e^{xln x}$.
    $endgroup$
    – Wuestenfux
    Nov 29 '18 at 15:14






  • 1




    $begingroup$
    $x=-1$ is an obvious solution.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:15












  • $begingroup$
    @AlexSilva Well, watch out for the RHS ...
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 15:24












  • $begingroup$
    $x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
    $endgroup$
    – Lance
    Nov 29 '18 at 15:27












  • $begingroup$
    @MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:30




















  • $begingroup$
    For $x>0$, $x^x = e^{xln x}$.
    $endgroup$
    – Wuestenfux
    Nov 29 '18 at 15:14






  • 1




    $begingroup$
    $x=-1$ is an obvious solution.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:15












  • $begingroup$
    @AlexSilva Well, watch out for the RHS ...
    $endgroup$
    – Michael Hoppe
    Nov 29 '18 at 15:24












  • $begingroup$
    $x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
    $endgroup$
    – Lance
    Nov 29 '18 at 15:27












  • $begingroup$
    @MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
    $endgroup$
    – Alex Silva
    Nov 29 '18 at 15:30


















$begingroup$
For $x>0$, $x^x = e^{xln x}$.
$endgroup$
– Wuestenfux
Nov 29 '18 at 15:14




$begingroup$
For $x>0$, $x^x = e^{xln x}$.
$endgroup$
– Wuestenfux
Nov 29 '18 at 15:14




1




1




$begingroup$
$x=-1$ is an obvious solution.
$endgroup$
– Alex Silva
Nov 29 '18 at 15:15






$begingroup$
$x=-1$ is an obvious solution.
$endgroup$
– Alex Silva
Nov 29 '18 at 15:15














$begingroup$
@AlexSilva Well, watch out for the RHS ...
$endgroup$
– Michael Hoppe
Nov 29 '18 at 15:24






$begingroup$
@AlexSilva Well, watch out for the RHS ...
$endgroup$
– Michael Hoppe
Nov 29 '18 at 15:24














$begingroup$
$x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
$endgroup$
– Lance
Nov 29 '18 at 15:27






$begingroup$
$x^x$ should be generally regarded as define on $(0,infty)$. Let $f(x)=x^x, g(x)=x^{-x}+(1+x)^{-x}$. Function graphs shows that there is one and only one intersection in $(1,2)$. Highly doubt there will be an analytical solution.
$endgroup$
– Lance
Nov 29 '18 at 15:27














$begingroup$
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
$endgroup$
– Alex Silva
Nov 29 '18 at 15:30






$begingroup$
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
$endgroup$
– Alex Silva
Nov 29 '18 at 15:30












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