Getting stuck on simple logarithmic equation: $x times ln (x) = 1$
$begingroup$
$$x times ln (x) = 1$$
I am trying to solve that equation. I used the theory $ln(a) = ln(b)$ being equivalent to $a = b$ and got stuck at
$$x = e^{frac{1}{x}}$$
That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.
logarithms
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
$endgroup$
add a comment |
$begingroup$
$$x times ln (x) = 1$$
I am trying to solve that equation. I used the theory $ln(a) = ln(b)$ being equivalent to $a = b$ and got stuck at
$$x = e^{frac{1}{x}}$$
That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.
logarithms
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
$endgroup$
$begingroup$
You'll probably have to use the Lambert W function.
$endgroup$
– Alex Becker
Apr 8 '13 at 21:32
$begingroup$
This is not an equation that can be solved with elementary functions, as far as I can tell.
$endgroup$
– Sammy Black
Apr 8 '13 at 21:34
add a comment |
$begingroup$
$$x times ln (x) = 1$$
I am trying to solve that equation. I used the theory $ln(a) = ln(b)$ being equivalent to $a = b$ and got stuck at
$$x = e^{frac{1}{x}}$$
That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.
logarithms
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
$endgroup$
$$x times ln (x) = 1$$
I am trying to solve that equation. I used the theory $ln(a) = ln(b)$ being equivalent to $a = b$ and got stuck at
$$x = e^{frac{1}{x}}$$
That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.
logarithms
logarithms
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
edited Dec 7 '18 at 12:26
Martin Sleziak
44.8k9118272
44.8k9118272
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
asked Apr 8 '13 at 21:28
David GomesDavid Gomes
122127
122127
$begingroup$
You'll probably have to use the Lambert W function.
$endgroup$
– Alex Becker
Apr 8 '13 at 21:32
$begingroup$
This is not an equation that can be solved with elementary functions, as far as I can tell.
$endgroup$
– Sammy Black
Apr 8 '13 at 21:34
add a comment |
$begingroup$
You'll probably have to use the Lambert W function.
$endgroup$
– Alex Becker
Apr 8 '13 at 21:32
$begingroup$
This is not an equation that can be solved with elementary functions, as far as I can tell.
$endgroup$
– Sammy Black
Apr 8 '13 at 21:34
$begingroup$
You'll probably have to use the Lambert W function.
$endgroup$
– Alex Becker
Apr 8 '13 at 21:32
$begingroup$
You'll probably have to use the Lambert W function.
$endgroup$
– Alex Becker
Apr 8 '13 at 21:32
$begingroup$
This is not an equation that can be solved with elementary functions, as far as I can tell.
$endgroup$
– Sammy Black
Apr 8 '13 at 21:34
$begingroup$
This is not an equation that can be solved with elementary functions, as far as I can tell.
$endgroup$
– Sammy Black
Apr 8 '13 at 21:34
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
You can use the law of logarithms which states that for $a,binmathbb{R}$: $aln{b}=lnleft(b^{a}right)$.
Therefore, you have:
$$xln{x}=1 implies ln{x^{x}}=1$$
You hence have:
$$x^{x}=e$$
Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $xapprox 1.76322$).
If you're interested, the closed form solution is: $$frac{1}{W(1)}, qquad text{ where } W(z) text{ is the LambertW function}$$
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
$endgroup$
add a comment |
$begingroup$
There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)
Basically, the Lambert-W function is the inverse function of:
$$f(x) = xe^x$$
Equivalently:
$$x=W(xe^x)$$
So, using your expression:
$$x=e^frac{1}{x}$$
$$1=frac{1}{x}e^frac{1}{x}$$
Taking the product-log of both sides:
$$W(1)=Wleft(frac{1}{x}e^frac{1}{x}right)$$
$$W(1)=frac{1}{x}$$
$$x=frac{1}{W(1)}$$
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
$endgroup$
add a comment |
$begingroup$
This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
$endgroup$
add a comment |
$begingroup$
It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation
$$x=L(x)e^{L(x)}$$
For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation
$$yln y=1$$
Then, the solution of your equation would be $ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
$endgroup$
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
add a comment |
$begingroup$
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
$endgroup$
1
$begingroup$
(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
$endgroup$
– apnorton
Apr 8 '13 at 21:38
$begingroup$
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
$endgroup$
– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can use the law of logarithms which states that for $a,binmathbb{R}$: $aln{b}=lnleft(b^{a}right)$.
Therefore, you have:
$$xln{x}=1 implies ln{x^{x}}=1$$
You hence have:
$$x^{x}=e$$
Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $xapprox 1.76322$).
If you're interested, the closed form solution is: $$frac{1}{W(1)}, qquad text{ where } W(z) text{ is the LambertW function}$$
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
$endgroup$
add a comment |
$begingroup$
You can use the law of logarithms which states that for $a,binmathbb{R}$: $aln{b}=lnleft(b^{a}right)$.
Therefore, you have:
$$xln{x}=1 implies ln{x^{x}}=1$$
You hence have:
$$x^{x}=e$$
Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $xapprox 1.76322$).
If you're interested, the closed form solution is: $$frac{1}{W(1)}, qquad text{ where } W(z) text{ is the LambertW function}$$
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
$endgroup$
add a comment |
$begingroup$
You can use the law of logarithms which states that for $a,binmathbb{R}$: $aln{b}=lnleft(b^{a}right)$.
Therefore, you have:
$$xln{x}=1 implies ln{x^{x}}=1$$
You hence have:
$$x^{x}=e$$
Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $xapprox 1.76322$).
If you're interested, the closed form solution is: $$frac{1}{W(1)}, qquad text{ where } W(z) text{ is the LambertW function}$$
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
$endgroup$
You can use the law of logarithms which states that for $a,binmathbb{R}$: $aln{b}=lnleft(b^{a}right)$.
Therefore, you have:
$$xln{x}=1 implies ln{x^{x}}=1$$
You hence have:
$$x^{x}=e$$
Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $xapprox 1.76322$).
If you're interested, the closed form solution is: $$frac{1}{W(1)}, qquad text{ where } W(z) text{ is the LambertW function}$$
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
answered Apr 8 '13 at 21:32
Thomas RussellThomas Russell
7,82632551
7,82632551
add a comment |
add a comment |
$begingroup$
There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)
Basically, the Lambert-W function is the inverse function of:
$$f(x) = xe^x$$
Equivalently:
$$x=W(xe^x)$$
So, using your expression:
$$x=e^frac{1}{x}$$
$$1=frac{1}{x}e^frac{1}{x}$$
Taking the product-log of both sides:
$$W(1)=Wleft(frac{1}{x}e^frac{1}{x}right)$$
$$W(1)=frac{1}{x}$$
$$x=frac{1}{W(1)}$$
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
$endgroup$
add a comment |
$begingroup$
There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)
Basically, the Lambert-W function is the inverse function of:
$$f(x) = xe^x$$
Equivalently:
$$x=W(xe^x)$$
So, using your expression:
$$x=e^frac{1}{x}$$
$$1=frac{1}{x}e^frac{1}{x}$$
Taking the product-log of both sides:
$$W(1)=Wleft(frac{1}{x}e^frac{1}{x}right)$$
$$W(1)=frac{1}{x}$$
$$x=frac{1}{W(1)}$$
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
$endgroup$
add a comment |
$begingroup$
There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)
Basically, the Lambert-W function is the inverse function of:
$$f(x) = xe^x$$
Equivalently:
$$x=W(xe^x)$$
So, using your expression:
$$x=e^frac{1}{x}$$
$$1=frac{1}{x}e^frac{1}{x}$$
Taking the product-log of both sides:
$$W(1)=Wleft(frac{1}{x}e^frac{1}{x}right)$$
$$W(1)=frac{1}{x}$$
$$x=frac{1}{W(1)}$$
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
$endgroup$
There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)
Basically, the Lambert-W function is the inverse function of:
$$f(x) = xe^x$$
Equivalently:
$$x=W(xe^x)$$
So, using your expression:
$$x=e^frac{1}{x}$$
$$1=frac{1}{x}e^frac{1}{x}$$
Taking the product-log of both sides:
$$W(1)=Wleft(frac{1}{x}e^frac{1}{x}right)$$
$$W(1)=frac{1}{x}$$
$$x=frac{1}{W(1)}$$
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
answered Apr 8 '13 at 21:37
apnortonapnorton
15.2k33796
15.2k33796
add a comment |
add a comment |
$begingroup$
This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
$endgroup$
add a comment |
$begingroup$
This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
$endgroup$
add a comment |
$begingroup$
This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
$endgroup$
This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
answered Apr 8 '13 at 21:31
icurays1icurays1
13.4k13155
13.4k13155
add a comment |
add a comment |
$begingroup$
It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation
$$x=L(x)e^{L(x)}$$
For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation
$$yln y=1$$
Then, the solution of your equation would be $ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
$endgroup$
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
add a comment |
$begingroup$
It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation
$$x=L(x)e^{L(x)}$$
For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation
$$yln y=1$$
Then, the solution of your equation would be $ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
$endgroup$
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
add a comment |
$begingroup$
It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation
$$x=L(x)e^{L(x)}$$
For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation
$$yln y=1$$
Then, the solution of your equation would be $ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
$endgroup$
It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation
$$x=L(x)e^{L(x)}$$
For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation
$$yln y=1$$
Then, the solution of your equation would be $ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
answered Apr 8 '13 at 21:50
paoloffpaoloff
262
262
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
add a comment |
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
$begingroup$
$L{}$ function?
$endgroup$
– apnorton
Apr 8 '13 at 21:52
add a comment |
$begingroup$
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
$endgroup$
1
$begingroup$
(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
$endgroup$
– apnorton
Apr 8 '13 at 21:38
$begingroup$
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
$endgroup$
– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
add a comment |
$begingroup$
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
$endgroup$
1
$begingroup$
(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
$endgroup$
– apnorton
Apr 8 '13 at 21:38
$begingroup$
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
$endgroup$
– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
add a comment |
$begingroup$
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
$endgroup$
${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
answered Apr 8 '13 at 21:32
Lord SothLord Soth
6,5901235
6,5901235
1
$begingroup$
(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
$endgroup$
– apnorton
Apr 8 '13 at 21:38
$begingroup$
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
$endgroup$
– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
add a comment |
1
$begingroup$
(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
$endgroup$
– apnorton
Apr 8 '13 at 21:38
$begingroup$
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
$endgroup$
– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
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– apnorton
Apr 8 '13 at 21:55
1
1
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(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
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– apnorton
Apr 8 '13 at 21:38
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(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though.
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– apnorton
Apr 8 '13 at 21:38
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@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
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– Lord Soth
Apr 8 '13 at 21:53
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@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here).
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– Lord Soth
Apr 8 '13 at 21:53
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
$begingroup$
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent.
$endgroup$
– apnorton
Apr 8 '13 at 21:55
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You'll probably have to use the Lambert W function.
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– Alex Becker
Apr 8 '13 at 21:32
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This is not an equation that can be solved with elementary functions, as far as I can tell.
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– Sammy Black
Apr 8 '13 at 21:34