How to solve $(x!)!+x!+x=x^x! $Solving equation involving factorialHow to solve this summation: $sum_k=1^nfrack2^k(k+2)!$Summation of a series help: $sum fracn-1n!$How to find the remainder when the following series is divided by 12?How to solve equation with factorial using algebra?Need help solving this inequality involving factorialHow to solve factorial equations with very big numbersHow do I calculate F if both sides have factorials?How to solve a binomial for $n$How do I solve $frac(n+2)!-n!5! = 330$?
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How to solve $(x!)!+x!+x=x^x! $
Solving equation involving factorialHow to solve this summation: $sum_k=1^nfrack2^k(k+2)!$Summation of a series help: $sum fracn-1n!$How to find the remainder when the following series is divided by 12?How to solve equation with factorial using algebra?Need help solving this inequality involving factorialHow to solve factorial equations with very big numbersHow do I calculate F if both sides have factorials?How to solve a binomial for $n$How do I solve $frac(n+2)!-n!5! = 330$?
$begingroup$
How to solve this equation
$$
(x!)!+x!+x=x^x!
$$
The answer is $3$ . But I have no idea of how to solve it. Thanks for your time.
factorial
edited 14 hours ago
dmtri
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How to solve this equation
$$
(x!)!+x!+x=x^x!
$$
The answer is $3$ . But I have no idea of how to solve it. Thanks for your time.
factorial
edited 14 hours ago
dmtri
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
8
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
4
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago
add a comment |
$begingroup$
How to solve this equation
$$
(x!)!+x!+x=x^x!
$$
The answer is $3$ . But I have no idea of how to solve it. Thanks for your time.
factorial
edited 14 hours ago
dmtri
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How to solve this equation
$$
(x!)!+x!+x=x^x!
$$
The answer is $3$ . But I have no idea of how to solve it. Thanks for your time.
factorial
factorial
edited 14 hours ago
dmtri
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
dmtri
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 14 hours ago
dmtri
1,5522521
edited 14 hours ago
dmtri
1,5522521
edited 14 hours ago
dmtri
1,5522521
1,5522521
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
asked 15 hours ago
Patric HornqvistPatric Hornqvist
312
312
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Patric Hornqvist is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
8
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
4
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago
add a comment |
1
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
8
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
4
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago
1
1
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
8
8
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
4
4
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I'm going to show $(x!)!>x^x!$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.
$(x!)!$ is a product of $x!$ factors, among which are $1,2,dots,x$ and $x^2+1,dots,x^2+x$ (since $x!geq x^2+x$ for $xgeq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore
$$(x!)!geq 1(x^2+1)cdot 2(x^2+2)cdotdotscdot x(x^2+x)cdot x^x!-2x>(x^2)^xcdot x^x!-2x=x^x!$$
as we wanted.
answered 14 hours ago
WojowuWojowu
18.8k23172
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I'm going to show $(x!)!>x^x!$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.
$(x!)!$ is a product of $x!$ factors, among which are $1,2,dots,x$ and $x^2+1,dots,x^2+x$ (since $x!geq x^2+x$ for $xgeq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore
$$(x!)!geq 1(x^2+1)cdot 2(x^2+2)cdotdotscdot x(x^2+x)cdot x^x!-2x>(x^2)^xcdot x^x!-2x=x^x!$$
as we wanted.
answered 14 hours ago
WojowuWojowu
18.8k23172
$endgroup$
add a comment |
$begingroup$
I'm going to show $(x!)!>x^x!$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.
$(x!)!$ is a product of $x!$ factors, among which are $1,2,dots,x$ and $x^2+1,dots,x^2+x$ (since $x!geq x^2+x$ for $xgeq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore
$$(x!)!geq 1(x^2+1)cdot 2(x^2+2)cdotdotscdot x(x^2+x)cdot x^x!-2x>(x^2)^xcdot x^x!-2x=x^x!$$
as we wanted.
answered 14 hours ago
WojowuWojowu
18.8k23172
$endgroup$
add a comment |
$begingroup$
I'm going to show $(x!)!>x^x!$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.
$(x!)!$ is a product of $x!$ factors, among which are $1,2,dots,x$ and $x^2+1,dots,x^2+x$ (since $x!geq x^2+x$ for $xgeq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore
$$(x!)!geq 1(x^2+1)cdot 2(x^2+2)cdotdotscdot x(x^2+x)cdot x^x!-2x>(x^2)^xcdot x^x!-2x=x^x!$$
as we wanted.
answered 14 hours ago
WojowuWojowu
18.8k23172
$endgroup$
I'm going to show $(x!)!>x^x!$ for $x>3$, which is clearly enough to show there are no solutions with $x>3$.
$(x!)!$ is a product of $x!$ factors, among which are $1,2,dots,x$ and $x^2+1,dots,x^2+x$ (since $x!geq x^2+x$ for $xgeq 4$), as well as $x!-2x$ other factors, each of which is at least $x$. Therefore
$$(x!)!geq 1(x^2+1)cdot 2(x^2+2)cdotdotscdot x(x^2+x)cdot x^x!-2x>(x^2)^xcdot x^x!-2x=x^x!$$
as we wanted.
answered 14 hours ago
WojowuWojowu
18.8k23172
answered 14 hours ago
WojowuWojowu
18.8k23172
answered 14 hours ago
WojowuWojowu
18.8k23172
answered 14 hours ago
WojowuWojowu
18.8k23172
18.8k23172
add a comment |
add a comment |
Patric Hornqvist is a new contributor. Be nice, and check out our Code of Conduct.
Patric Hornqvist is a new contributor. Be nice, and check out our Code of Conduct.
Patric Hornqvist is a new contributor. Be nice, and check out our Code of Conduct.
Patric Hornqvist is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Are you solving in the real numbers?
$endgroup$
– Toby Mak
15 hours ago
8
$begingroup$
Are you trying to prove that $x=3$ is the only answer?
$endgroup$
– Barry Cipra
15 hours ago
4
$begingroup$
$x approx 2.282$ is also a solution if one replaces factorial by gamma function
$endgroup$
– YuiTo Cheng
15 hours ago