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

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

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.

• substitute $x_i=exp(b_i)$.
– YJT
Aug 20, 2020 at 12:36
• It would be better if you wrote the whole problem out using latex with MathJax, rather than relying on a picture. This makes your question easier for others to edit. For example in the first three lines the letter "o" should be the number $0$. See: math.meta.stackexchange.com/questions/5020/… Aug 20, 2020 at 12:40
• YJT's hint looks good to me. Use that and the problem should become much more manageable. Aug 20, 2020 at 12:50

## 1 Answer

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.