Mixing
Calculating multiple parameters for a logarithmic function
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
I have the following function -
$$y=ln(Ax^D+B+Cx^E)$$
These are the coordinates:
$$(1, 1)$$$$(2, 0.84)$$$$(4, 1.5)$$$$(31, 4.1)$$$$(44, 5)$$
How can you solve this equation and what is the correct method for finding the $A,B,C,D$ and $E$ parameters?
EDIT
It seems that the last coordinate is causing a problem.
Anyway what's the correct way to find the equation without the last coordinate and as close as possible?
calculus algebra-precalculus functions logarithms systems-of-equations
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
asked Jul 15 at 8:34
Drxxd
1305
 |Â
show 15 more comments
up vote
2
down vote
favorite
I have the following function -
$$y=ln(Ax^D+B+Cx^E)$$
These are the coordinates:
$$(1, 1)$$$$(2, 0.84)$$$$(4, 1.5)$$$$(31, 4.1)$$$$(44, 5)$$
How can you solve this equation and what is the correct method for finding the $A,B,C,D$ and $E$ parameters?
EDIT
It seems that the last coordinate is causing a problem.
Anyway what's the correct way to find the equation without the last coordinate and as close as possible?
calculus algebra-precalculus functions logarithms systems-of-equations
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
asked Jul 15 at 8:34
Drxxd
1305
2
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
1
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
1
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
1
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57
 |Â
show 15 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I have the following function -
$$y=ln(Ax^D+B+Cx^E)$$
These are the coordinates:
$$(1, 1)$$$$(2, 0.84)$$$$(4, 1.5)$$$$(31, 4.1)$$$$(44, 5)$$
How can you solve this equation and what is the correct method for finding the $A,B,C,D$ and $E$ parameters?
EDIT
It seems that the last coordinate is causing a problem.
Anyway what's the correct way to find the equation without the last coordinate and as close as possible?
calculus algebra-precalculus functions logarithms systems-of-equations
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
asked Jul 15 at 8:34
Drxxd
1305
I have the following function -
$$y=ln(Ax^D+B+Cx^E)$$
These are the coordinates:
$$(1, 1)$$$$(2, 0.84)$$$$(4, 1.5)$$$$(31, 4.1)$$$$(44, 5)$$
How can you solve this equation and what is the correct method for finding the $A,B,C,D$ and $E$ parameters?
EDIT
It seems that the last coordinate is causing a problem.
Anyway what's the correct way to find the equation without the last coordinate and as close as possible?
calculus algebra-precalculus functions logarithms systems-of-equations
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
asked Jul 15 at 8:34
Drxxd
1305
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
edited Jul 15 at 9:07
TheSimpliFire
9,69261951
9,69261951
asked Jul 15 at 8:34
Drxxd
1305
asked Jul 15 at 8:34
Drxxd
1305
asked Jul 15 at 8:34
Drxxd
1305
1305
2
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
1
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
1
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
1
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57
 |Â
show 15 more comments
2
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
1
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
1
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
1
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57
2
2
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
1
1
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
1
1
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
1
1
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57
 |Â
show 15 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Too long for a comment, but may be of some use:
I have omitted the last set of coordinates as the OP has requested.
Substitute $(1,1)$: $$A+B+C=etag1$$
Substitute $(2,0.84)$: $$Acdot2^D+Ccdot2^E+B=e^0.84tag2$$
Substitute $(4,1.5)$: $$Acdot2^2D+Ccdot2^2E+B=e^4tag3$$
Substitute $(31,4.1)$: $$Acdot31^D+Ccdot31^E+B=e^4.1tag4$$
Do $(3)-(2)$: $$Acdot2^D(2^D-1)+Ccdot2^E(2^E-1)=e^4-e^0.84tag5$$
Do $(2)-(1)$: $$Acdot(2^D-1)+Ccdot(2^E-1)=e^0.84-etag6$$
Simultaneous equation so solve for $A$ and $C$ in terms of $D$ and $E$.
Multiply $(7)$ by $2^E$: $$Acdot2^E(2^D-1)+Ccdot2^E(2^E-1)=2^E(e^4-e^0.84)tag7$$
Do $(8)-(6)$: $$A(2^E-2^D)(2^D-1)=(2^E-1)(e^4-e^0.84)$$ so $$A=frac(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag8$$
Substitute $(9)$ into $(7)$: $$C=frac(2^E-2^D)(e^0.84-e)-(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag9$$
Substitute $(9)$ and $(10)$ into $(1)$: $$B=e-frace^0.84-e2^D-1tag10$$
Now $A$ and $C$ are in terms of $D$ and $E$, but $B$ is only in terms of $D$.
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
 |Â
show 4 more comments
up vote
0
down vote
The model
$$Ax^D+B+Cx^E=e^y$$ is partly linear and partly nonlinear and has no closed-form solution. You can make it slightly more tractable numerically by plugging the coordinates of the five points to obtain five equations, and eliminate the coefficients $A,B,C$ by forming linear combinations (for instance, you can solve the three first equations by Cramer and plug in the remaining two).
In the end, you will obtain as simplified system of two equations in the two unknowns $D$ and $E$ (though these equations involve ten exponentials). This can be solved by Newton in a less costly way than the original system, and as there are only two independent variables left, graphical observation is possible.
answered Jul 15 at 9:38
Yves Daoust
111k665204
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Too long for a comment, but may be of some use:
I have omitted the last set of coordinates as the OP has requested.
Substitute $(1,1)$: $$A+B+C=etag1$$
Substitute $(2,0.84)$: $$Acdot2^D+Ccdot2^E+B=e^0.84tag2$$
Substitute $(4,1.5)$: $$Acdot2^2D+Ccdot2^2E+B=e^4tag3$$
Substitute $(31,4.1)$: $$Acdot31^D+Ccdot31^E+B=e^4.1tag4$$
Do $(3)-(2)$: $$Acdot2^D(2^D-1)+Ccdot2^E(2^E-1)=e^4-e^0.84tag5$$
Do $(2)-(1)$: $$Acdot(2^D-1)+Ccdot(2^E-1)=e^0.84-etag6$$
Simultaneous equation so solve for $A$ and $C$ in terms of $D$ and $E$.
Multiply $(7)$ by $2^E$: $$Acdot2^E(2^D-1)+Ccdot2^E(2^E-1)=2^E(e^4-e^0.84)tag7$$
Do $(8)-(6)$: $$A(2^E-2^D)(2^D-1)=(2^E-1)(e^4-e^0.84)$$ so $$A=frac(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag8$$
Substitute $(9)$ into $(7)$: $$C=frac(2^E-2^D)(e^0.84-e)-(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag9$$
Substitute $(9)$ and $(10)$ into $(1)$: $$B=e-frace^0.84-e2^D-1tag10$$
Now $A$ and $C$ are in terms of $D$ and $E$, but $B$ is only in terms of $D$.
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
 |Â
show 4 more comments
up vote
1
down vote
accepted
Too long for a comment, but may be of some use:
I have omitted the last set of coordinates as the OP has requested.
Substitute $(1,1)$: $$A+B+C=etag1$$
Substitute $(2,0.84)$: $$Acdot2^D+Ccdot2^E+B=e^0.84tag2$$
Substitute $(4,1.5)$: $$Acdot2^2D+Ccdot2^2E+B=e^4tag3$$
Substitute $(31,4.1)$: $$Acdot31^D+Ccdot31^E+B=e^4.1tag4$$
Do $(3)-(2)$: $$Acdot2^D(2^D-1)+Ccdot2^E(2^E-1)=e^4-e^0.84tag5$$
Do $(2)-(1)$: $$Acdot(2^D-1)+Ccdot(2^E-1)=e^0.84-etag6$$
Simultaneous equation so solve for $A$ and $C$ in terms of $D$ and $E$.
Multiply $(7)$ by $2^E$: $$Acdot2^E(2^D-1)+Ccdot2^E(2^E-1)=2^E(e^4-e^0.84)tag7$$
Do $(8)-(6)$: $$A(2^E-2^D)(2^D-1)=(2^E-1)(e^4-e^0.84)$$ so $$A=frac(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag8$$
Substitute $(9)$ into $(7)$: $$C=frac(2^E-2^D)(e^0.84-e)-(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag9$$
Substitute $(9)$ and $(10)$ into $(1)$: $$B=e-frace^0.84-e2^D-1tag10$$
Now $A$ and $C$ are in terms of $D$ and $E$, but $B$ is only in terms of $D$.
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Too long for a comment, but may be of some use:
I have omitted the last set of coordinates as the OP has requested.
Substitute $(1,1)$: $$A+B+C=etag1$$
Substitute $(2,0.84)$: $$Acdot2^D+Ccdot2^E+B=e^0.84tag2$$
Substitute $(4,1.5)$: $$Acdot2^2D+Ccdot2^2E+B=e^4tag3$$
Substitute $(31,4.1)$: $$Acdot31^D+Ccdot31^E+B=e^4.1tag4$$
Do $(3)-(2)$: $$Acdot2^D(2^D-1)+Ccdot2^E(2^E-1)=e^4-e^0.84tag5$$
Do $(2)-(1)$: $$Acdot(2^D-1)+Ccdot(2^E-1)=e^0.84-etag6$$
Simultaneous equation so solve for $A$ and $C$ in terms of $D$ and $E$.
Multiply $(7)$ by $2^E$: $$Acdot2^E(2^D-1)+Ccdot2^E(2^E-1)=2^E(e^4-e^0.84)tag7$$
Do $(8)-(6)$: $$A(2^E-2^D)(2^D-1)=(2^E-1)(e^4-e^0.84)$$ so $$A=frac(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag8$$
Substitute $(9)$ into $(7)$: $$C=frac(2^E-2^D)(e^0.84-e)-(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag9$$
Substitute $(9)$ and $(10)$ into $(1)$: $$B=e-frace^0.84-e2^D-1tag10$$
Now $A$ and $C$ are in terms of $D$ and $E$, but $B$ is only in terms of $D$.
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
Too long for a comment, but may be of some use:
I have omitted the last set of coordinates as the OP has requested.
Substitute $(1,1)$: $$A+B+C=etag1$$
Substitute $(2,0.84)$: $$Acdot2^D+Ccdot2^E+B=e^0.84tag2$$
Substitute $(4,1.5)$: $$Acdot2^2D+Ccdot2^2E+B=e^4tag3$$
Substitute $(31,4.1)$: $$Acdot31^D+Ccdot31^E+B=e^4.1tag4$$
Do $(3)-(2)$: $$Acdot2^D(2^D-1)+Ccdot2^E(2^E-1)=e^4-e^0.84tag5$$
Do $(2)-(1)$: $$Acdot(2^D-1)+Ccdot(2^E-1)=e^0.84-etag6$$
Simultaneous equation so solve for $A$ and $C$ in terms of $D$ and $E$.
Multiply $(7)$ by $2^E$: $$Acdot2^E(2^D-1)+Ccdot2^E(2^E-1)=2^E(e^4-e^0.84)tag7$$
Do $(8)-(6)$: $$A(2^E-2^D)(2^D-1)=(2^E-1)(e^4-e^0.84)$$ so $$A=frac(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag8$$
Substitute $(9)$ into $(7)$: $$C=frac(2^E-2^D)(e^0.84-e)-(2^E-1)(e^4-e^0.84)(2^E-2^D)(2^D-1)tag9$$
Substitute $(9)$ and $(10)$ into $(1)$: $$B=e-frace^0.84-e2^D-1tag10$$
Now $A$ and $C$ are in terms of $D$ and $E$, but $B$ is only in terms of $D$.
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
edited Jul 15 at 9:04
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
answered Jul 15 at 8:59
TheSimpliFire
9,69261951
9,69261951
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
 |Â
show 4 more comments
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
If we had another equation then the expressions for $A,B,C$ could be substituted into the remaining two equations and could be solved simultaneously.
– TheSimpliFire
Jul 15 at 9:10
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
That's great for this moment.. Thanks!
– Drxxd
Jul 15 at 9:18
1
1
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
No, $4^D=(2^2)^D=2^2Dneq2^D+1$
– TheSimpliFire
Jul 16 at 10:43
1
1
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
the concept is still the same whether it is 1.5 or 4
– TheSimpliFire
Jul 16 at 10:43
1
1
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
Are you sure? Try $D=5$...
– TheSimpliFire
Jul 16 at 13:18
 |Â
show 4 more comments
up vote
0
down vote
The model
$$Ax^D+B+Cx^E=e^y$$ is partly linear and partly nonlinear and has no closed-form solution. You can make it slightly more tractable numerically by plugging the coordinates of the five points to obtain five equations, and eliminate the coefficients $A,B,C$ by forming linear combinations (for instance, you can solve the three first equations by Cramer and plug in the remaining two).
In the end, you will obtain as simplified system of two equations in the two unknowns $D$ and $E$ (though these equations involve ten exponentials). This can be solved by Newton in a less costly way than the original system, and as there are only two independent variables left, graphical observation is possible.
answered Jul 15 at 9:38
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
The model
$$Ax^D+B+Cx^E=e^y$$ is partly linear and partly nonlinear and has no closed-form solution. You can make it slightly more tractable numerically by plugging the coordinates of the five points to obtain five equations, and eliminate the coefficients $A,B,C$ by forming linear combinations (for instance, you can solve the three first equations by Cramer and plug in the remaining two).
In the end, you will obtain as simplified system of two equations in the two unknowns $D$ and $E$ (though these equations involve ten exponentials). This can be solved by Newton in a less costly way than the original system, and as there are only two independent variables left, graphical observation is possible.
answered Jul 15 at 9:38
Yves Daoust
111k665204
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The model
$$Ax^D+B+Cx^E=e^y$$ is partly linear and partly nonlinear and has no closed-form solution. You can make it slightly more tractable numerically by plugging the coordinates of the five points to obtain five equations, and eliminate the coefficients $A,B,C$ by forming linear combinations (for instance, you can solve the three first equations by Cramer and plug in the remaining two).
In the end, you will obtain as simplified system of two equations in the two unknowns $D$ and $E$ (though these equations involve ten exponentials). This can be solved by Newton in a less costly way than the original system, and as there are only two independent variables left, graphical observation is possible.
answered Jul 15 at 9:38
Yves Daoust
111k665204
The model
$$Ax^D+B+Cx^E=e^y$$ is partly linear and partly nonlinear and has no closed-form solution. You can make it slightly more tractable numerically by plugging the coordinates of the five points to obtain five equations, and eliminate the coefficients $A,B,C$ by forming linear combinations (for instance, you can solve the three first equations by Cramer and plug in the remaining two).
In the end, you will obtain as simplified system of two equations in the two unknowns $D$ and $E$ (though these equations involve ten exponentials). This can be solved by Newton in a less costly way than the original system, and as there are only two independent variables left, graphical observation is possible.
answered Jul 15 at 9:38
Yves Daoust
111k665204
edited Jul 15 at 9:47
answered Jul 15 at 9:38
Yves Daoust
111k665204
answered Jul 15 at 9:38
Yves Daoust
111k665204
answered Jul 15 at 9:38
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852303%2fcalculating-multiple-parameters-for-a-logarithmic-function%23new-answer', 'question_page');
);
Post as a guest
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
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
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
2
desmos.com/calculator/ragn6n1qqr
– Mason
Jul 15 at 8:38
I've also tried to play with desmos but I couldn't get the exact function, so the question is what's the correct way to solve this
– Drxxd
Jul 15 at 8:41
1
Where did you get this problem from?
– TheSimpliFire
Jul 15 at 8:47
1
@Mason fitting the points in semi-log scale gives $R^2 = 1$ even with the fifth point desmos.com/calculator/vm1eh3qb28 sorry for the link before, I just realised that I gave your link instead of mine
– Davide Morgante
Jul 15 at 9:04
1
@Mason ahah don't worry! It wasn't my brightest moment, got wrong too many things! My line of reasoning wanted to be the same as the answer of@YvesDaoust
– Davide Morgante
Jul 15 at 13:57