# Consider the two equations below: y 1 </msub> = ( 1 &#x22

Consider the two equations below:
${y}_{1}=\left(1-\frac{{a}_{1}}{x}\right){e}^{-\frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{1}}{x}}\phantom{\rule{0ex}{0ex}}{y}_{2}=\left(1-\frac{{a}_{2}}{x}\right){e}^{-\frac{\alpha \phantom{\rule{thinmathspace}{0ex}}{a}_{2}}{x}}$
Given ${y}_{1}$, ${y}_{2}$, ${a}_{1}$ and ${a}_{2}$, is there an analytical way to determine $\alpha$ and $x$?
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sniokd
A plot of the function reveals a solution close to x=0.01 and the solution, obtained using Newton, is x=0.0133409. From here, alpha=1.584997.
I also used the solution based on the Lambert function. Here again, the plot of the fuction reveal a root close to alpha=1.5. Starting from here, Newton iterations lead again to alpha=1.584997. Frm here, x=0.0133409.
Then, again, both approaches work and lead to the same results.
Please post where are your problems. As told earlier, the problem is a resonable starting guess for the solution; this can be easily obtained looking at the plot of the function.
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letumsnemesislh
Note that
$\frac{{y}_{1}^{{a}_{2}}}{{y}_{2}^{{a}_{1}}}=\frac{{\left(1-\frac{{a}_{1}}{x}\right)}^{{a}_{2}}}{{\left(1-\frac{{a}_{2}}{x}\right)}^{{a}_{1}}}$
This eliminates $\alpha$.
Not sure how you can solve this analytically. But once you get $x$, you can solve for $\alpha$ by taking logarithm of either equation.