Exponential Equation Calculus












0












$begingroup$


$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.



What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $



Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $



Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
    $endgroup$
    – Viktor Glombik
    Dec 24 '18 at 0:31










  • $begingroup$
    I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – rafa11111
    Dec 24 '18 at 0:33










  • $begingroup$
    Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
    $endgroup$
    – David K
    Dec 24 '18 at 1:09










  • $begingroup$
    @David K I updated. Thank you.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:46










  • $begingroup$
    @raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:54
















0












$begingroup$


$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.



What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $



Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $



Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.






calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
    $endgroup$
    – Viktor Glombik
    Dec 24 '18 at 0:31










  • $begingroup$
    I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – rafa11111
    Dec 24 '18 at 0:33










  • $begingroup$
    Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
    $endgroup$
    – David K
    Dec 24 '18 at 1:09










  • $begingroup$
    @David K I updated. Thank you.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:46










  • $begingroup$
    @raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:54














0












0








0





$begingroup$


$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.



What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $



Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $



Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.






calculus







share|cite|improve this question











$endgroup$




$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.



What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $



Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $



Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 16:36







JDragon314159

















asked Dec 24 '18 at 0:11









JDragon314159JDragon314159

137




137












  • $begingroup$
    What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
    $endgroup$
    – Viktor Glombik
    Dec 24 '18 at 0:31










  • $begingroup$
    I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – rafa11111
    Dec 24 '18 at 0:33










  • $begingroup$
    Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
    $endgroup$
    – David K
    Dec 24 '18 at 1:09










  • $begingroup$
    @David K I updated. Thank you.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:46










  • $begingroup$
    @raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:54


















  • $begingroup$
    What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
    $endgroup$
    – Viktor Glombik
    Dec 24 '18 at 0:31










  • $begingroup$
    I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
    $endgroup$
    – rafa11111
    Dec 24 '18 at 0:33










  • $begingroup$
    Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
    $endgroup$
    – David K
    Dec 24 '18 at 1:09










  • $begingroup$
    @David K I updated. Thank you.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:46










  • $begingroup$
    @raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 1:54
















$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31




$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31












$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33




$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33












$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09




$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09












$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46




$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46












$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54




$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54










1 Answer
1






active

oldest

votes


















0












$begingroup$

All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$



In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.



If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.



If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.



If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .



If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .



If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .



This allows to answer to your questions I) and II).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 13:46












  • $begingroup$
    The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:36












  • $begingroup$
    Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:51












  • $begingroup$
    I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
    $endgroup$
    – JDragon314159
    Dec 27 '18 at 21:30










  • $begingroup$
    What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
    $endgroup$
    – JJacquelin
    Dec 28 '18 at 6:30













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









0












$begingroup$

All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$



In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.



If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.



If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.



If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .



If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .



If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .



This allows to answer to your questions I) and II).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 13:46












  • $begingroup$
    The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:36












  • $begingroup$
    Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:51












  • $begingroup$
    I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
    $endgroup$
    – JDragon314159
    Dec 27 '18 at 21:30










  • $begingroup$
    What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
    $endgroup$
    – JJacquelin
    Dec 28 '18 at 6:30


















0












$begingroup$

All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$



In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.



If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.



If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.



If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .



If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .



If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .



This allows to answer to your questions I) and II).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 13:46












  • $begingroup$
    The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:36












  • $begingroup$
    Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:51












  • $begingroup$
    I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
    $endgroup$
    – JDragon314159
    Dec 27 '18 at 21:30










  • $begingroup$
    What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
    $endgroup$
    – JJacquelin
    Dec 28 '18 at 6:30
















0












0








0





$begingroup$

All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$



In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.



If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.



If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.



If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .



If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .



If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .



This allows to answer to your questions I) and II).






share|cite|improve this answer











$endgroup$



All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$



In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.



If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.



If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.



If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .



If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .



If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .



This allows to answer to your questions I) and II).







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edited Dec 24 '18 at 9:06

























answered Dec 24 '18 at 7:58









JJacquelinJJacquelin

43.7k21853




43.7k21853












  • $begingroup$
    I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 13:46












  • $begingroup$
    The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:36












  • $begingroup$
    Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:51












  • $begingroup$
    I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
    $endgroup$
    – JDragon314159
    Dec 27 '18 at 21:30










  • $begingroup$
    What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
    $endgroup$
    – JJacquelin
    Dec 28 '18 at 6:30




















  • $begingroup$
    I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
    $endgroup$
    – JDragon314159
    Dec 24 '18 at 13:46












  • $begingroup$
    The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:36












  • $begingroup$
    Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
    $endgroup$
    – JJacquelin
    Dec 24 '18 at 16:51












  • $begingroup$
    I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
    $endgroup$
    – JDragon314159
    Dec 27 '18 at 21:30










  • $begingroup$
    What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
    $endgroup$
    – JJacquelin
    Dec 28 '18 at 6:30


















$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46






$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46














$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36






$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36














$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51






$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51














$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30




$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30












$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30






$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30




















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