Mixing
Algebra and exponential functions
$begingroup$
Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$.
Thanks.
logarithms
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
$endgroup$
add a comment |
$begingroup$
Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$.
Thanks.
logarithms
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
$endgroup$
$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01
add a comment |
$begingroup$
Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$.
Thanks.
logarithms
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
$endgroup$
Given $x^n=y$ where $n=y$, I have no problem finding $x$ if $y$ is known. Problem is getting $y$ when only the value of $x$ is known. Is there a way? At the moment I'm working with $x$ and $y$ values less than $1$.
Thanks.
logarithms
logarithms
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
edited Feb 3 at 15:00
J. W. Tanner
4,5391320
4,5391320
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
asked Jan 31 at 17:58
Adam BoitAdam Boit
164
164
$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01
add a comment |
$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01
$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y implies x^{-y} = y^{-1} implies yx^{-y}=1 => ye^{-ln(x)y} =1$
Multiply by $-ln x$ on both sides:
$-yln xe^{-yln x}=-ln x$
Now use the Lambert W function on both sides.
$-yln x=W(-ln x) implies y=-frac {W(-ln x)}{ln x}$
For $x lt 1$, $W(-ln x)$ is well defined, so this is our solution.
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
|
show 3 more comments
$begingroup$
If $$x^n=y$$ so we get $$x=sqrt[n]{y}$$
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y implies x^{-y} = y^{-1} implies yx^{-y}=1 => ye^{-ln(x)y} =1$
Multiply by $-ln x$ on both sides:
$-yln xe^{-yln x}=-ln x$
Now use the Lambert W function on both sides.
$-yln x=W(-ln x) implies y=-frac {W(-ln x)}{ln x}$
For $x lt 1$, $W(-ln x)$ is well defined, so this is our solution.
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
|
show 3 more comments
$begingroup$
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y implies x^{-y} = y^{-1} implies yx^{-y}=1 => ye^{-ln(x)y} =1$
Multiply by $-ln x$ on both sides:
$-yln xe^{-yln x}=-ln x$
Now use the Lambert W function on both sides.
$-yln x=W(-ln x) implies y=-frac {W(-ln x)}{ln x}$
For $x lt 1$, $W(-ln x)$ is well defined, so this is our solution.
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
|
show 3 more comments
$begingroup$
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y implies x^{-y} = y^{-1} implies yx^{-y}=1 => ye^{-ln(x)y} =1$
Multiply by $-ln x$ on both sides:
$-yln xe^{-yln x}=-ln x$
Now use the Lambert W function on both sides.
$-yln x=W(-ln x) implies y=-frac {W(-ln x)}{ln x}$
For $x lt 1$, $W(-ln x)$ is well defined, so this is our solution.
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
$endgroup$
So you are trying to solve for $y$ in the equation $x^y = y$.
The Lambert W function is useful for this.
$x^y=y implies x^{-y} = y^{-1} implies yx^{-y}=1 => ye^{-ln(x)y} =1$
Multiply by $-ln x$ on both sides:
$-yln xe^{-yln x}=-ln x$
Now use the Lambert W function on both sides.
$-yln x=W(-ln x) implies y=-frac {W(-ln x)}{ln x}$
For $x lt 1$, $W(-ln x)$ is well defined, so this is our solution.
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
edited Jan 31 at 18:18
Mohammad Zuhair Khan
1,6912625
1,6912625
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
answered Jan 31 at 18:04
Abhishek VangipuramAbhishek Vangipuram
694
694
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
|
show 3 more comments
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
Wow! I would never have found that on my own. I struggled for the past couple of days in vain. A couple of hours ago I got a hunch (epiphany?) Natural logarithm is involved somehow. Thanks a lot.
$endgroup$
– Adam Boit
Jan 31 at 18:46
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
I have found a way that does not require Natural logarithm. It is repeated iteration with Common Logarithm (decidic). Is that new or it's an obsolete method?
$endgroup$
– Adam Boit
Feb 1 at 13:51
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
can you share your function for y in terms of x (y(x))?
$endgroup$
– Abhishek Vangipuram
Feb 2 at 23:59
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Sure. x^y=y means log to base x of y is y. Therefore [math]log(y)/y=log(x)[/math]
$endgroup$
– Adam Boit
Feb 3 at 11:58
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
$begingroup$
Yes, this is trivial. However it does not help you find y for a given x like my solution.
$endgroup$
– Abhishek Vangipuram
Feb 3 at 19:52
|
show 3 more comments
$begingroup$
If $$x^n=y$$ so we get $$x=sqrt[n]{y}$$
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
$endgroup$
add a comment |
$begingroup$
If $$x^n=y$$ so we get $$x=sqrt[n]{y}$$
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
$endgroup$
add a comment |
$begingroup$
If $$x^n=y$$ so we get $$x=sqrt[n]{y}$$
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
$endgroup$
If $$x^n=y$$ so we get $$x=sqrt[n]{y}$$
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
answered Jan 31 at 18:00
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.7k42867
78.7k42867
add a comment |
add a comment |
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$begingroup$
In the elementary functions it's impossible.
$endgroup$
– Michael Rozenberg
Jan 31 at 18:00
$begingroup$
$ln(x^y)=yln(x)$
$endgroup$
– hamam_Abdallah
Jan 31 at 18:01