How to solve a system of linear equation with $e$.

Question

$$0.449=\frac{e^{b_1}}{e^{b_1}+e^{b_2}+e^{b_3}}$$ $$0.335=\frac{e^{b_2}}{e^{b_1}+e^{b_2}+e^{b_3}}$$ $$0.216=\frac{e^{b_3}}{e^{b_1}+e^{b_2}+e^{b_3}}$$

My attempt: $$\ln(0.449)=b_1+\ln(.)$$ $$\ln(0.335)=b_2+\ln(.)$$ $$\ln(0.216)=b_3+\ln(.)$$ where $\ln(.)=\ln(e^{b_1}+e^{b_2}+e^{b_3})$

But then I get stuck and would like help on how to solve this? What are the values for $b_1$, $b_2$ and $b_3$?

Would like a step-by-step explanation. Thanks.

Answer 1

There are infinitely many solutions. To prove this, let $A=e^{b_1}, B=e^{b_2}, C=e^{b_3}$. $A,B$ and $C$ all must be positive numbers. Then the system of equations becomes: $$0.449=\frac{A}{A+B+C}$$ $$0.335=\frac{B}{A+B+C}$$ $$0.216=\frac{C}{A+B+C}$$

There are infinitely many solutions, for which we have one degree of freedom. Our solutions are: $$A=0.449D\Rightarrow b_1=\ln(0.449)+\ln(D) = \ln(0.449) + E$$ $$B=0.335D\Rightarrow b_2=\ln(0.335)+\ln(D) = \ln(0.335) + E$$ $$C=0.216D\Rightarrow b_3=\ln(0.216)+\ln(D) = \ln(0.216) + E$$ $D$ can take on any positive value, and $E=\ln(D)$ can take on any real value.