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Are there other solutions of $x^x^3-x=2^x^2+x$ than $x=-1 $ and $x=2$ in $mathbbR$?
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I have tried to solve that equation $x^x^3-x=2^x^2+x$ in $mathbbR$ , I have got only two integers solutions which they are : $x=-1$, $x=2$ , are there others ?
Note: if we try to study this : $fracx^3-xx^2+x=fraclog xlog 2$, I think there is a numerical solution in the range $( 0,1)$ using value intermediate theorem, it is to show that there is a solution here, but i can't determine it
real-analysis lambert-w
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
 |Â
show 2 more comments
up vote
0
down vote
favorite
I have tried to solve that equation $x^x^3-x=2^x^2+x$ in $mathbbR$ , I have got only two integers solutions which they are : $x=-1$, $x=2$ , are there others ?
Note: if we try to study this : $fracx^3-xx^2+x=fraclog xlog 2$, I think there is a numerical solution in the range $( 0,1)$ using value intermediate theorem, it is to show that there is a solution here, but i can't determine it
real-analysis lambert-w
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
1
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
1
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
1
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have tried to solve that equation $x^x^3-x=2^x^2+x$ in $mathbbR$ , I have got only two integers solutions which they are : $x=-1$, $x=2$ , are there others ?
Note: if we try to study this : $fracx^3-xx^2+x=fraclog xlog 2$, I think there is a numerical solution in the range $( 0,1)$ using value intermediate theorem, it is to show that there is a solution here, but i can't determine it
real-analysis lambert-w
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
I have tried to solve that equation $x^x^3-x=2^x^2+x$ in $mathbbR$ , I have got only two integers solutions which they are : $x=-1$, $x=2$ , are there others ?
Note: if we try to study this : $fracx^3-xx^2+x=fraclog xlog 2$, I think there is a numerical solution in the range $( 0,1)$ using value intermediate theorem, it is to show that there is a solution here, but i can't determine it
real-analysis lambert-w
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
asked Jul 20 at 20:33
zeraoulia rafik
2,1231823
2,1231823
1
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
1
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
1
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37
 |Â
show 2 more comments
1
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
1
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
1
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37
1
1
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
1
1
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
1
1
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
0
down vote
HINT: Write your equation in the form
$$ln(x)-fracx^2+xx^3-1ln(2)=0$$, define
$$f(x)=ln(x)-fracx^2+xx^3-1ln(2)$$ and use calculus.
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
add a comment |Â
up vote
0
down vote
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)log 2= log xcdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $log 2= log xcdot (x-1)$, or equivalently $log 2- log xcdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= log 2- log xcdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
answered Jul 20 at 21:23
A. Pongrácz
2,319221
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT: Write your equation in the form
$$ln(x)-fracx^2+xx^3-1ln(2)=0$$, define
$$f(x)=ln(x)-fracx^2+xx^3-1ln(2)$$ and use calculus.
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
add a comment |Â
up vote
0
down vote
HINT: Write your equation in the form
$$ln(x)-fracx^2+xx^3-1ln(2)=0$$, define
$$f(x)=ln(x)-fracx^2+xx^3-1ln(2)$$ and use calculus.
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT: Write your equation in the form
$$ln(x)-fracx^2+xx^3-1ln(2)=0$$, define
$$f(x)=ln(x)-fracx^2+xx^3-1ln(2)$$ and use calculus.
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
HINT: Write your equation in the form
$$ln(x)-fracx^2+xx^3-1ln(2)=0$$, define
$$f(x)=ln(x)-fracx^2+xx^3-1ln(2)$$ and use calculus.
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
answered Jul 20 at 21:03
Dr. Sonnhard Graubner
66.8k32659
66.8k32659
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
add a comment |Â
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
So you want to tell me that you have a closed form for the $0.346323dots$ solution?
– asdf
Jul 20 at 21:11
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
This can i not tell you, sorry, but you can construct the image of the function and you will get the Information, how many solution the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:14
1
1
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
@Dr.Sonnhard Graubner , But you didn't add any thing to my above note ,I want to add me somthing otherwise than i have wrote above
– zeraoulia rafik
Jul 20 at 21:19
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
You will never find all solutions of your equation, but with my hint, you can prove how many solutions the equation has.
– Dr. Sonnhard Graubner
Jul 20 at 21:24
add a comment |Â
up vote
0
down vote
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)log 2= log xcdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $log 2= log xcdot (x-1)$, or equivalently $log 2- log xcdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= log 2- log xcdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
answered Jul 20 at 21:23
A. Pongrácz
2,319221
add a comment |Â
up vote
0
down vote
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)log 2= log xcdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $log 2= log xcdot (x-1)$, or equivalently $log 2- log xcdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= log 2- log xcdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
answered Jul 20 at 21:23
A. Pongrácz
2,319221
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)log 2= log xcdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $log 2= log xcdot (x-1)$, or equivalently $log 2- log xcdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= log 2- log xcdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
answered Jul 20 at 21:23
A. Pongrácz
2,319221
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)log 2= log xcdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $log 2= log xcdot (x-1)$, or equivalently $log 2- log xcdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= log 2- log xcdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
answered Jul 20 at 21:23
A. Pongrácz
2,319221
edited Jul 20 at 21:36
answered Jul 20 at 21:23
A. Pongrácz
2,319221
answered Jul 20 at 21:23
A. Pongrácz
2,319221
answered Jul 20 at 21:23
A. Pongrácz
2,319221
2,319221
add a comment |Â
add a comment |Â
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1
Another obvious solution is $x=0$ because $0^0=1=2^0$
– Hagen von Eitzen
Jul 20 at 20:35
Wolfram finds a solution around $x≈0.346323$.
– lulu
Jul 20 at 20:35
Thanks so much, i didn't montioned that because i have consedred x=0 as indeterminate case
– zeraoulia rafik
Jul 20 at 20:36
1
I don't think we should consider 0 a solution as $0^0=1$ is debatable.
– Shrey Joshi
Jul 20 at 20:36
1
$0^0$ is an indeterminate form (mostly because $(x,y)mapsto x^y$ is not continuous there) but not an undefined expression. $0^0$ is the number of maps from the empty set to the empty set.
– Hagen von Eitzen
Jul 20 at 20:37