I was looking for inspirations for solving the below equation for x $$ -e^x \ln \left( \frac{(e^x -2 \alpha)(1+\alpha)}{1-\alpha} \right) + xe^x +2\alpha e^x - 4 \alpha^2 - 2\alpha = 0$$ where $0<\alpha<1$.

Is is analytically possible at all?

Many thanks.

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    $\begingroup$ no. almost certainly not. $\endgroup$ – Bennett Gardiner Jul 6 '15 at 0:13
  • $\begingroup$ Thanks! That's also helpful. $\endgroup$ – Learner Jul 6 '15 at 1:15
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    $\begingroup$ Equations which mix polynomial, logarithmic and exponential functions do not show explicit solutions and numerical method should be used. $\endgroup$ – Claude Leibovici Jul 6 '15 at 4:53

As said, this equation will not show any explicit solution.

However, you could notice that the equation $$f(x,a)=-e^x \ln \left( \frac{(e^x -2 a)(1+a)}{1-a} \right) + xe^x +2a e^x - 4 a^2 - 2a = 0 $$ has two roots in $x$ for given $a$ and one of them (apparently the largest one) seems to be close to $a$ (just smaller than $a$). Expanding $f(x,a)$ as a Taylor series built at $x=a$, $$f(a,a)\approx \frac{4 a^4}{3}-\frac{2 a^5}{3}+O\left(a^6\right)$$ So, Newton method starting at $x_0=a$ would quickly converge to the root.

  • $\begingroup$ Thanks Cluade. It's good to know and helpful in giving an approximation to the solution. $\endgroup$ – Learner Jul 7 '15 at 6:02

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