solving for x in $x^x-x^{-x}=(1+x)^{-x}$
1
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
edited Nov 29 at 15:18
axblount
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
|
show 2 more comments
1
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
edited Nov 29 at 15:18
axblount
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
1
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
$x^x$ 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.
– Lance
Nov 29 at 15:27
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30
|
show 2 more comments
1
1
1
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
edited Nov 29 at 15:18
axblount
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
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
edited Nov 29 at 15:18
axblount
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
edited Nov 29 at 15:18
axblount
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
edited Nov 29 at 15:18
axblount
1,99811016
edited Nov 29 at 15:18
axblount
1,99811016
edited Nov 29 at 15:18
axblount
1,99811016
1,99811016
asked Nov 29 at 15:01
Juan Pablo Arcila
412
asked Nov 29 at 15:01
Juan Pablo Arcila
412
asked Nov 29 at 15:01
Juan Pablo Arcila
412
412
For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
1
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
$x^x$ 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.
– Lance
Nov 29 at 15:27
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30
|
show 2 more comments
For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
1
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
$x^x$ 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.
– Lance
Nov 29 at 15:27
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30
For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
1
1
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
$x^x$ 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.
– Lance
Nov 29 at 15:27
$x^x$ 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.
– Lance
Nov 29 at 15:27
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30
|
show 2 more comments
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For $x>0$, $x^x = e^{xln x}$.
– Wuestenfux
Nov 29 at 15:14
1
$x=-1$ is an obvious solution.
– Alex Silva
Nov 29 at 15:15
@AlexSilva Well, watch out for the RHS ...
– Michael Hoppe
Nov 29 at 15:24
$x^x$ 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.
– Lance
Nov 29 at 15:27
@MichaelHoppe, the RHS gives $(1-1)^{1}=0$. Am I wrong? I think $x^x$ is well defined for $x < 0$ and $x in mathbb{Z}$.
– Alex Silva
Nov 29 at 15:30