solving for a variable that exist inside as well as outside of natural log or exponent
up vote
0
down vote
favorite
can the following equation be solved for K analytically? If not, then what other approaches I could try out?
K*ln[(C2-K)/(C1-K)] = -(F/V)*t
The original equation was:
C2 = K + (C1-K)*exp(-(F/KV)*t)
logarithms exponential-function
asked Jan 25 '16 at 1:34
ToNoY
13315
add a comment |
up vote
0
down vote
favorite
can the following equation be solved for K analytically? If not, then what other approaches I could try out?
K*ln[(C2-K)/(C1-K)] = -(F/V)*t
The original equation was:
C2 = K + (C1-K)*exp(-(F/KV)*t)
logarithms exponential-function
asked Jan 25 '16 at 1:34
ToNoY
13315
1
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
can the following equation be solved for K analytically? If not, then what other approaches I could try out?
K*ln[(C2-K)/(C1-K)] = -(F/V)*t
The original equation was:
C2 = K + (C1-K)*exp(-(F/KV)*t)
logarithms exponential-function
asked Jan 25 '16 at 1:34
ToNoY
13315
can the following equation be solved for K analytically? If not, then what other approaches I could try out?
K*ln[(C2-K)/(C1-K)] = -(F/V)*t
The original equation was:
C2 = K + (C1-K)*exp(-(F/KV)*t)
logarithms exponential-function
logarithms exponential-function
asked Jan 25 '16 at 1:34
ToNoY
13315
asked Jan 25 '16 at 1:34
ToNoY
13315
edited Jan 25 '16 at 2:03
asked Jan 25 '16 at 1:34
ToNoY
13315
asked Jan 25 '16 at 1:34
ToNoY
13315
asked Jan 25 '16 at 1:34
ToNoY
13315
13315
1
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50
add a comment |
1
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50
1
1
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
To start, I point out a few things.
A) You could simply solve for $K$ using Lagrange Inversion Theorem.
B) You could try to solve for $K$ in closed form using the Lambert W function.
As for A), you will need to understand calculus.
As for B), I know it won't work, it only works to solve some problems of this type.
I can solve the following:
$$Kln(K)=A$$
$$K^K=e^A$$
$$K=e^{W(A)}$$
Solution here.
However, a problem like:
$$Kln(K+a)=A$$
Is unsolvable.
Yours is also unsolvable.
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
To start, I point out a few things.
A) You could simply solve for $K$ using Lagrange Inversion Theorem.
B) You could try to solve for $K$ in closed form using the Lambert W function.
As for A), you will need to understand calculus.
As for B), I know it won't work, it only works to solve some problems of this type.
I can solve the following:
$$Kln(K)=A$$
$$K^K=e^A$$
$$K=e^{W(A)}$$
Solution here.
However, a problem like:
$$Kln(K+a)=A$$
Is unsolvable.
Yours is also unsolvable.
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
add a comment |
up vote
0
down vote
To start, I point out a few things.
A) You could simply solve for $K$ using Lagrange Inversion Theorem.
B) You could try to solve for $K$ in closed form using the Lambert W function.
As for A), you will need to understand calculus.
As for B), I know it won't work, it only works to solve some problems of this type.
I can solve the following:
$$Kln(K)=A$$
$$K^K=e^A$$
$$K=e^{W(A)}$$
Solution here.
However, a problem like:
$$Kln(K+a)=A$$
Is unsolvable.
Yours is also unsolvable.
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
add a comment |
up vote
0
down vote
up vote
0
down vote
To start, I point out a few things.
A) You could simply solve for $K$ using Lagrange Inversion Theorem.
B) You could try to solve for $K$ in closed form using the Lambert W function.
As for A), you will need to understand calculus.
As for B), I know it won't work, it only works to solve some problems of this type.
I can solve the following:
$$Kln(K)=A$$
$$K^K=e^A$$
$$K=e^{W(A)}$$
Solution here.
However, a problem like:
$$Kln(K+a)=A$$
Is unsolvable.
Yours is also unsolvable.
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
To start, I point out a few things.
A) You could simply solve for $K$ using Lagrange Inversion Theorem.
B) You could try to solve for $K$ in closed form using the Lambert W function.
As for A), you will need to understand calculus.
As for B), I know it won't work, it only works to solve some problems of this type.
I can solve the following:
$$Kln(K)=A$$
$$K^K=e^A$$
$$K=e^{W(A)}$$
Solution here.
However, a problem like:
$$Kln(K+a)=A$$
Is unsolvable.
Yours is also unsolvable.
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
answered Jan 27 '16 at 22:56
Simply Beautiful Art
50.1k577181
50.1k577181
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1625783%2fsolving-for-a-variable-that-exist-inside-as-well-as-outside-of-natural-log-or-ex%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Do not expect this to have a nice solution. Look into root-finding (en.wikipedia.org/wiki/Root-finding_algorithm) instead.
– parsiad
Jan 25 '16 at 2:13
You would have to use the Lambert W function to find it in closed terms.
– Simply Beautiful Art
Jan 27 '16 at 22:49
You can call $-F/V*t$ another variable so that it doesn't look as messy.
– Simply Beautiful Art
Jan 27 '16 at 22:50