What is J0?
WolframAlpha finds a solution, but the implicit plot is not really useful.
Dear Sirs/Madams.
I'm trying to solve numerically this equation:
K1= (Jp/J0) * ( (J0-B) / (Jp-B) )
K2= ( (1/Jp) - (1/J0) )
t= (1/(A*B^2)) * ln ( K1 - B * K2) <- this is the equation
Where t, J0, A and B are constants
and Jp is unknowned.
In my example:
t=0.1
A=1,88E+05
B=6,974
When I try to solve numérically (bisection method and secant method),
log of a negative value doesn't exist and then there is an error.
I have transformed the equation into another one, taking exponential function left and right
and then the value of the exponential is a very big number
and the method fails too (note that A is a big number).
I have tried with Matlab.
I'm sure the solution exists because
I have had the A an B values using
a mathemathical software (MathCad).
Does anyone have a solution
or an idea to solve this problem ?
I will be very gratefull to you.
Thanks in advance.
Pepelu.
What is J0?
WolframAlpha finds a solution, but the implicit plot is not really useful.
Thank you 'mfb' for your answer!
I forgot to mention that J0 is another constant.
However, I think that the equation in WolframAlpha is not the same that I wrote.
Inside the log there must be (K1-B K2).
In addition, WolframAlpha does not tell how it is solved.
The equation will be:
0.1 = 1/ (1.88*10^5 * 6.974^2) * ln ( ((x/0.9)* ((0.9-6.974) / (x-6.974))) - 6.974* ((1/x)-(1/0.9 )))
The Wolfram solution is wrong: -1.627 e^-397105
The correct answer (I know it) is arround J=0.9
Any other idea ?
I think the board, WolframAlpha or my browser messed the link up. It came from the correct equation.
earboths solutions look good: The argument of the logarithm has to be very large, this can be done in two ways: Large 1/(Jp-B) in K1 or large 1/Jp in K2. Those options lead to the posted solutions.
Something close to 0.9 cannot solve the equation, the argument of the logarithm would be too small.