# Solving a exponential/log equation

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.

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

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.