How to findout this equations value
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John D'Errico on 5 Jul 2022
Edited: John D'Errico on 5 Jul 2022
That is impossible, since your equations are not written in unambiguous mathematics.
Is e^15b intended to mean exp(15*b)?
Is e^15b intended to mean exp(15)*b?
Is e the base of the natural logarithem? Thus
Even if as I suspect, you actually have these equations:
syms R b
E1 = R*exp(15*b) == 35.75;
E2 = R*exp(25*b) == 29.125;
E3 = R*exp(35*b) == 22.875;
then no unique solution can possibly exist. That is, divide equation 2 by equation 1. That leaves us with
As you can see, R goes away, and we could compute b from that relation.
However, suppose we chose to do this instead?
And finally, we could do this:
Which as you see, returns a completely different result for b. And when we do so, depending on the value of b we chose from how we might derive it, then R is also completely different.
More Answers (2)
Saksham Gupta on 5 Jul 2022
As per my understanding, you wish to solve the given equations
You may use the following code to solve them:
syms R b
eqns = [ R*exp(15*b) == 35.75 , R*exp(25*b) == 29.125 ];
a = solve( eqns , [ R ,b ] );
I am using 2 equations only as I am able to identify only 2 variables : 'R' and 'b'.
My output is:
You may refer to documentation of solve to learn more about it.
Sam Chak on 5 Jul 2022
You can try using the solve() command.
R12 = 48.6173; % solving Eqn 1 and Eqn 2 simultaneously
R23 = 53.2756; % solving Eqn 2 and Eqn 3 simultaneously
R31 = 49.9703; % solving Eqn 3 and Eqn 1 simultaneously
b12 = -0.020495335725415202423043826631973699411349908259696363801512111;
b23 = -0.024155230072427962515559053432074702762580354955690539307109413;
b31 = -0.022325282898921582469301440032024201086965131607693451554310762;
x = linspace(10, 40, 3001);
y1 = R12*exp(b12*x);
y2 = R23*exp(b23*x);
y3 = R31*exp(b31*x);
plot(x', [y1' y2' y3'], 'LineWidth', 1.5), grid on, hold on,
plot([15 25 35], [35.75 29.125 22.875], 'mo', 'MarkerSize', 12, 'LineWidth', 2), hold off
legend('y1', 'y2', 'y3')