Solution for $x = a^x$
$begingroup$
We know $e^x = x$ has no real solution, and $1^x = x$ has a solution $x=1$.
For what value $1 < a < e$, $a^x = x$ has a real solution? (and no $b>a$ yields a solution.)
I do not have any context for the question, become curious when I used $x leq e^x$.
Appreciate any comment.
elementary-functions
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
$endgroup$
add a comment |
$begingroup$
We know $e^x = x$ has no real solution, and $1^x = x$ has a solution $x=1$.
For what value $1 < a < e$, $a^x = x$ has a real solution? (and no $b>a$ yields a solution.)
I do not have any context for the question, become curious when I used $x leq e^x$.
Appreciate any comment.
elementary-functions
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
$endgroup$
add a comment |
$begingroup$
We know $e^x = x$ has no real solution, and $1^x = x$ has a solution $x=1$.
For what value $1 < a < e$, $a^x = x$ has a real solution? (and no $b>a$ yields a solution.)
I do not have any context for the question, become curious when I used $x leq e^x$.
Appreciate any comment.
elementary-functions
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
$endgroup$
We know $e^x = x$ has no real solution, and $1^x = x$ has a solution $x=1$.
For what value $1 < a < e$, $a^x = x$ has a real solution? (and no $b>a$ yields a solution.)
I do not have any context for the question, become curious when I used $x leq e^x$.
Appreciate any comment.
elementary-functions
elementary-functions
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
asked Dec 22 '18 at 9:41
Jo'Jo'
306110
306110
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Taking logs of both sides of $e^{log(a)x} = a^x = x$ gives $log(a)x = log(x)$, and rearranging gives $frac{log(x)}{x} = log(a)$. The function $frac{log(x)}{x}$ takes values between $-infty$ and $frac{1}{e}$ (this is not too hard to check). Therefore, $a^x = x$ has a solution for any $0 < a < e^{frac{1}{e}}$.
In fact, we can do better: $a^x = x$ has exactly one solution for $0 < a leq 1$ and $a = e^{frac{1}{e}}$, and exactly two solutions for $1 < a < e^{frac{1}{e}}$. This can again be deduced by looking at the graph of $frac{log(x)}{x}$: 
answered Dec 22 '18 at 9:49
ODFODF
1,486510
$endgroup$
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
add a comment |
$begingroup$
This equation can be solved using a function known as Lambert $W$. It's the inverse of $f(x)=x e^x$
https://en.m.wikipedia.org/wiki/Lambert_W_function
Take your equation and write it as $e^{x log (a)} =x$, then $-xlog( a) e^{-x log (a)}= -log( a)$, apply $W$ to get $-x log( a)= W(-log (a))$, so $x = -W(-log (a))/log (a)$.
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
oldest
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active
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votes
$begingroup$
Taking logs of both sides of $e^{log(a)x} = a^x = x$ gives $log(a)x = log(x)$, and rearranging gives $frac{log(x)}{x} = log(a)$. The function $frac{log(x)}{x}$ takes values between $-infty$ and $frac{1}{e}$ (this is not too hard to check). Therefore, $a^x = x$ has a solution for any $0 < a < e^{frac{1}{e}}$.
In fact, we can do better: $a^x = x$ has exactly one solution for $0 < a leq 1$ and $a = e^{frac{1}{e}}$, and exactly two solutions for $1 < a < e^{frac{1}{e}}$. This can again be deduced by looking at the graph of $frac{log(x)}{x}$:
answered Dec 22 '18 at 9:49
ODFODF
1,486510
$endgroup$
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
add a comment |
$begingroup$
Taking logs of both sides of $e^{log(a)x} = a^x = x$ gives $log(a)x = log(x)$, and rearranging gives $frac{log(x)}{x} = log(a)$. The function $frac{log(x)}{x}$ takes values between $-infty$ and $frac{1}{e}$ (this is not too hard to check). Therefore, $a^x = x$ has a solution for any $0 < a < e^{frac{1}{e}}$.
In fact, we can do better: $a^x = x$ has exactly one solution for $0 < a leq 1$ and $a = e^{frac{1}{e}}$, and exactly two solutions for $1 < a < e^{frac{1}{e}}$. This can again be deduced by looking at the graph of $frac{log(x)}{x}$:
answered Dec 22 '18 at 9:49
ODFODF
1,486510
$endgroup$
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
add a comment |
$begingroup$
Taking logs of both sides of $e^{log(a)x} = a^x = x$ gives $log(a)x = log(x)$, and rearranging gives $frac{log(x)}{x} = log(a)$. The function $frac{log(x)}{x}$ takes values between $-infty$ and $frac{1}{e}$ (this is not too hard to check). Therefore, $a^x = x$ has a solution for any $0 < a < e^{frac{1}{e}}$.
In fact, we can do better: $a^x = x$ has exactly one solution for $0 < a leq 1$ and $a = e^{frac{1}{e}}$, and exactly two solutions for $1 < a < e^{frac{1}{e}}$. This can again be deduced by looking at the graph of $frac{log(x)}{x}$:
answered Dec 22 '18 at 9:49
ODFODF
1,486510
$endgroup$
Taking logs of both sides of $e^{log(a)x} = a^x = x$ gives $log(a)x = log(x)$, and rearranging gives $frac{log(x)}{x} = log(a)$. The function $frac{log(x)}{x}$ takes values between $-infty$ and $frac{1}{e}$ (this is not too hard to check). Therefore, $a^x = x$ has a solution for any $0 < a < e^{frac{1}{e}}$.
In fact, we can do better: $a^x = x$ has exactly one solution for $0 < a leq 1$ and $a = e^{frac{1}{e}}$, and exactly two solutions for $1 < a < e^{frac{1}{e}}$. This can again be deduced by looking at the graph of $frac{log(x)}{x}$:
answered Dec 22 '18 at 9:49
ODFODF
1,486510
answered Dec 22 '18 at 9:49
ODFODF
1,486510
answered Dec 22 '18 at 9:49
ODFODF
1,486510
answered Dec 22 '18 at 9:49
ODFODF
1,486510
1,486510
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
add a comment |
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
Many thanks, will accept in 2 min.
$endgroup$
– Jo'
Dec 22 '18 at 9:53
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
$begingroup$
And again only one solution at the boundary $a=e^{frac{1}{e}}$
$endgroup$
– Todor Markov
Dec 22 '18 at 10:09
add a comment |
$begingroup$
This equation can be solved using a function known as Lambert $W$. It's the inverse of $f(x)=x e^x$
https://en.m.wikipedia.org/wiki/Lambert_W_function
Take your equation and write it as $e^{x log (a)} =x$, then $-xlog( a) e^{-x log (a)}= -log( a)$, apply $W$ to get $-x log( a)= W(-log (a))$, so $x = -W(-log (a))/log (a)$.
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
$endgroup$
add a comment |
$begingroup$
This equation can be solved using a function known as Lambert $W$. It's the inverse of $f(x)=x e^x$
https://en.m.wikipedia.org/wiki/Lambert_W_function
Take your equation and write it as $e^{x log (a)} =x$, then $-xlog( a) e^{-x log (a)}= -log( a)$, apply $W$ to get $-x log( a)= W(-log (a))$, so $x = -W(-log (a))/log (a)$.
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
$endgroup$
add a comment |
$begingroup$
This equation can be solved using a function known as Lambert $W$. It's the inverse of $f(x)=x e^x$
https://en.m.wikipedia.org/wiki/Lambert_W_function
Take your equation and write it as $e^{x log (a)} =x$, then $-xlog( a) e^{-x log (a)}= -log( a)$, apply $W$ to get $-x log( a)= W(-log (a))$, so $x = -W(-log (a))/log (a)$.
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
$endgroup$
This equation can be solved using a function known as Lambert $W$. It's the inverse of $f(x)=x e^x$
https://en.m.wikipedia.org/wiki/Lambert_W_function
Take your equation and write it as $e^{x log (a)} =x$, then $-xlog( a) e^{-x log (a)}= -log( a)$, apply $W$ to get $-x log( a)= W(-log (a))$, so $x = -W(-log (a))/log (a)$.
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
answered Dec 22 '18 at 10:56
jjagmathjjagmath
2577
2577
add a comment |
add a comment |
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