Is my solution to $a^x=x+b$ correct?
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I was trying to find a general solution to the equation $a^x=x+b$. First, I used a substitution:
$$u=a^xLongleftrightarrow x=log_au$$
Then, it went as follows:
$$u=log_a u+b$$
$$-b=log_a u-u$$
$$-b=log_a u-log_a a^u$$
$$-b=log_a ua^-u$$
$$a^-b=ua^-u$$
$$-a^-b=-ua^-u$$
$$-u=Wbig(!-!a^-b,big)$$
$$u=-Wbig(!-!a^-b,big)$$
$$a^x=-Wbig(!-!a^-b,big)$$
I finally got the solution of:
$$x=lnBig(!-Wbig(!-!a^-b,big)Big)overln a$$
The problem:
From a paper on the Lambert W function, it says that a general solution to the same equation is:
$$x=-b-Wbig(!-!a^-b;ln a,big)overln a$$
which is quite different from what I got. I thought that maybe it is the same thing, but I wasn't able to manipulate my answer to get to the paper's answer.
So, the question is, how do I prove or disprove that:
$$lnBig(!-Wbig(!-!a^-b,big)Big)overln a=-b-Wbig(!-!a^-b;ln a,big)overln a$$
EDIT: As Ahmed noted in the comments, I made an error in my work, I mistaked $-ua^-u$ for $-ue^-u$, and incorrectly took the Lambert W function of it, instead of changing the exp base first. When I corrected myself, I managed to arrive at the correct answer.
exponential-function lambert-w
asked Sep 8 at 5:49
KKZiomek
1,6561134
 |Â
show 1 more comment
up vote
2
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I was trying to find a general solution to the equation $a^x=x+b$. First, I used a substitution:
$$u=a^xLongleftrightarrow x=log_au$$
Then, it went as follows:
$$u=log_a u+b$$
$$-b=log_a u-u$$
$$-b=log_a u-log_a a^u$$
$$-b=log_a ua^-u$$
$$a^-b=ua^-u$$
$$-a^-b=-ua^-u$$
$$-u=Wbig(!-!a^-b,big)$$
$$u=-Wbig(!-!a^-b,big)$$
$$a^x=-Wbig(!-!a^-b,big)$$
I finally got the solution of:
$$x=lnBig(!-Wbig(!-!a^-b,big)Big)overln a$$
The problem:
From a paper on the Lambert W function, it says that a general solution to the same equation is:
$$x=-b-Wbig(!-!a^-b;ln a,big)overln a$$
which is quite different from what I got. I thought that maybe it is the same thing, but I wasn't able to manipulate my answer to get to the paper's answer.
So, the question is, how do I prove or disprove that:
$$lnBig(!-Wbig(!-!a^-b,big)Big)overln a=-b-Wbig(!-!a^-b;ln a,big)overln a$$
EDIT: As Ahmed noted in the comments, I made an error in my work, I mistaked $-ua^-u$ for $-ue^-u$, and incorrectly took the Lambert W function of it, instead of changing the exp base first. When I corrected myself, I managed to arrive at the correct answer.
exponential-function lambert-w
asked Sep 8 at 5:49
KKZiomek
1,6561134
How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I was trying to find a general solution to the equation $a^x=x+b$. First, I used a substitution:
$$u=a^xLongleftrightarrow x=log_au$$
Then, it went as follows:
$$u=log_a u+b$$
$$-b=log_a u-u$$
$$-b=log_a u-log_a a^u$$
$$-b=log_a ua^-u$$
$$a^-b=ua^-u$$
$$-a^-b=-ua^-u$$
$$-u=Wbig(!-!a^-b,big)$$
$$u=-Wbig(!-!a^-b,big)$$
$$a^x=-Wbig(!-!a^-b,big)$$
I finally got the solution of:
$$x=lnBig(!-Wbig(!-!a^-b,big)Big)overln a$$
The problem:
From a paper on the Lambert W function, it says that a general solution to the same equation is:
$$x=-b-Wbig(!-!a^-b;ln a,big)overln a$$
which is quite different from what I got. I thought that maybe it is the same thing, but I wasn't able to manipulate my answer to get to the paper's answer.
So, the question is, how do I prove or disprove that:
$$lnBig(!-Wbig(!-!a^-b,big)Big)overln a=-b-Wbig(!-!a^-b;ln a,big)overln a$$
EDIT: As Ahmed noted in the comments, I made an error in my work, I mistaked $-ua^-u$ for $-ue^-u$, and incorrectly took the Lambert W function of it, instead of changing the exp base first. When I corrected myself, I managed to arrive at the correct answer.
exponential-function lambert-w
asked Sep 8 at 5:49
KKZiomek
1,6561134
I was trying to find a general solution to the equation $a^x=x+b$. First, I used a substitution:
$$u=a^xLongleftrightarrow x=log_au$$
Then, it went as follows:
$$u=log_a u+b$$
$$-b=log_a u-u$$
$$-b=log_a u-log_a a^u$$
$$-b=log_a ua^-u$$
$$a^-b=ua^-u$$
$$-a^-b=-ua^-u$$
$$-u=Wbig(!-!a^-b,big)$$
$$u=-Wbig(!-!a^-b,big)$$
$$a^x=-Wbig(!-!a^-b,big)$$
I finally got the solution of:
$$x=lnBig(!-Wbig(!-!a^-b,big)Big)overln a$$
The problem:
From a paper on the Lambert W function, it says that a general solution to the same equation is:
$$x=-b-Wbig(!-!a^-b;ln a,big)overln a$$
which is quite different from what I got. I thought that maybe it is the same thing, but I wasn't able to manipulate my answer to get to the paper's answer.
So, the question is, how do I prove or disprove that:
$$lnBig(!-Wbig(!-!a^-b,big)Big)overln a=-b-Wbig(!-!a^-b;ln a,big)overln a$$
EDIT: As Ahmed noted in the comments, I made an error in my work, I mistaked $-ua^-u$ for $-ue^-u$, and incorrectly took the Lambert W function of it, instead of changing the exp base first. When I corrected myself, I managed to arrive at the correct answer.
exponential-function lambert-w
exponential-function lambert-w
asked Sep 8 at 5:49
KKZiomek
1,6561134
asked Sep 8 at 5:49
KKZiomek
1,6561134
edited Sep 8 at 6:18
asked Sep 8 at 5:49
KKZiomek
1,6561134
asked Sep 8 at 5:49
KKZiomek
1,6561134
asked Sep 8 at 5:49
KKZiomek
1,6561134
1,6561134
How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16
 |Â
show 1 more comment
How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16
How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16
 |Â
show 1 more comment
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How’d you go from $-ua^-u=-a^-b$ to $-u=W(-a^b)$. What if $a neq e$? The inverse of $xcolorrede^x$ is $W(x)$.
– Ahmed S. Attaalla
Sep 8 at 5:53
Oh! I'm so stupid, that's what happens when I do math in the middle of the night. Thank you, so it seems my answer is incorrect.
– KKZiomek
Sep 8 at 5:54
No you’re not :). We all make mistakes...
– Ahmed S. Attaalla
Sep 8 at 5:56
I don't know where you found the solution. To me, it should be $$x=-b-fracWleft(-a^-b log (a)right)log (a)$$ Typo's ?
– Claude Leibovici
Sep 8 at 6:06
@ClaudeLeibovici yes that's a typo, I'll fix my question. And also btw, I did it correctly now and arrived at the correct answer. Thanks to Ahmed for spotting my error :)
– KKZiomek
Sep 8 at 6:16