I'm trying to solve this equation (analytically)

$e^{2x^2+3x-1}$ - $3e^{x^2+1}$ + $5e^{2x+3}=0$

By Descartes rule of signs I know there are no more than 4 roots. With two terms it is easy to convert into a polynomial but three (or more) exponential terms is a challenge.

Tried to factor out common terms but it leads to another similar equation. A numerical solution is easy so my interest is in an algebraic method.

I have searched for similar but no luck.

All hints, pointers and solutions appreciated.


They are not indices but exponents.

There is no way to find any analytical solution of this highly transcendental equation and numerical methods will be required.

You could consider that you look for the zero's of function $$f(x)=e^{2 x^2+3 x-1}-3 e^{x^2+1}+5 e^{2 x+3}$$ but it so stiff that is is difficult so see where are, more or less, the solutions.

Rearrange it, take logarithms and consider function $$g(x)=\log \left(e^{2 x^2+3 x-1}+5 e^{2 x+3}\right)-(x^2+1)\log \left(3\right)$$ which is reallmuch nicer. If you oberlap the plots ot the two logarithms, the problem "looks" like the intersetion of two parabolas showing solutions close to $-4$ and $-1$.

Now, we are ready for a use of Newton method which, using these initial values, will generate the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -1.000000 \\ 1 & -0.879291 \\ 2 & -0.875603 \\ 3 & -0.875599 \end{array} \right)$$

$$\left( \begin{array}{cc} n & x_n \\ 0 & -4.000000 \\ 1 & -3.819722 \\ 2 & -3.812717 \\ 3 & -3.812707 \end{array} \right)$$

  • $\begingroup$ Thanks but I managed to find a way to turn it into a quadratic in $(e^{x})^x$ and then factorise and solve each part on my calc. $\endgroup$ – Quadratic Reciprocity Oct 5 '20 at 23:54

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