I am trying to solve this implicit equation

[ccdean100]

Apologies, there were some mistakes in that equation. Here is the correct one:


(A*((D*ECARB)+(E*log((F+(G*ECARB))/H)))) = 1

Again, for ECARB

Can anyone help with this?

Charles

 

[Deleted member 4993]

Apologies, there were some mistakes in that equation. Here is the correct one:


(A*((D*ECARB)+(E*log((F+(G*ECARB))/H)))) = 1

Again, for ECARB

Can anyone help with this?

Charles

substitute

ECARB = x

(A*((D*ECARB)+(E*log((F+(G*ECARB))/H)))) = 1

A*[D*x+E*log{(F+G*x)/H}] = 1

D*x + E*log{(F+G*x)/H} = 1/A

This is a non-linear equation - do you know Newton_Raphson method?

 

[mmm4444bot]

I'm not sure what an "implicit equation" is.

I put your equation into Maple (assuming that the logarithm is base-10), and it reported a solution in terms of the LambertW function.

[attachment=0:th0sgxl5]lambertw.JPG[/attachment:th0sgxl5]

 

[ccdean100]

Hi, I've heard of it but never used it. Is there are straightforward way of using the method for this example?

 

[Deleted member 4993]

Hi, I've heard of it but never used it. Is there are straightforward way of using the method for this example?

Yes - if you know the numerical values of those constants (A, G, etc.)

 

[ccdean100]

Yes, all of the constants are known... any tips on how to use the Netwon method in this example?

 

[ccdean100]

Re:

I'm not sure what an "implicit equation" is.

I put your equation into Maple (assuming that the logarithm is base-10), and it reported a solution in terms of the LambertW function.

[attachment=0:1a2bzivd]lambertw.JPG[/attachment:1a2bzivd]

Thankyou, but in the equation you show, at the end, there is "ln(10)" ("EG+FDln(10)") but there is nothing inside of the log term, so what does that mean in Maple speak?! - it appears to be natural log of something, but of what?

Many thanks.

 

[Deleted member 4993]

Yes, all of the constants are known... any tips on how to use the Netwon method in this example?

Use Google - read up on the topic - tell us what you found and how do you propose to use the method.

 

[Deleted member 4993]

ln(10) = 2.302585093

 

[mmm4444bot]

My version of Maple likes natural logarithms; it usually converts base-10 to base-e (without asking) using the change-of-base formula:

\(\displaystyle \log_{10}(x) \;=\; \frac{ln(x)}{ln(10)}\)