When I try to solve this exponential equation:
Solve[E^-t^2 (2 - E^-t^2)+E^(-2 t - (3 t^2)/2) (2 - E^-t^2) (2 - E^(-2 t - t^2/2)) (1-E^-t^2 (2 - E^-t^2)) == 0.1, t, Reals]
I don't get any solution despite waiting for a long time
How can I solve it and is there another way Please Help
I would do the following. First of all, define the auxiliary function $f(t)$ as follows
f[t_]:=E^-t^2 (2 - E^-t^2)+E^(-2 t - (3 t^2)/2) (2 - E^-t^2) (2 - E^(-2 t - t^2/2)) (1-E^-t^2 (2 - E^-t^2))-0.1
Thus, the equation you want to solve is $f(t)=0$. The plot of $f(t)$ may indicate if there are solutions to this equation. Indeed, use
Plot[f[t], {t, -1, 2}]
We explicitly see that one solution is roughly located at $x\approx-0.8$ and the second one at $x\approx 1.5$. This piece of information is useful.
Now, by means of FindRoot one can obtain the roots using those rough estimates to establish an appropriate domain to look for the root. For example, the negative root is contained in [-1,0]. Thus,
FindRoot[f[t] == 0, {t, -1, 0}]
{t -> -0.751553}
Analogously, for the positive root. Of course you need to check whether there are more roots in an extended domain and not only inside [-1,2].
Solve[E^-t^2 (2 - E^-t^2) + E^(-2 t - (3 t^2)/2) (2 - E^-t^2) (2 - E^(-2 t - t^2/2)) (1 - E^-t^2 (2 - E^-t^2)) == 1/10 && -10 < t < 10, t, Reals]– Mariusz Iwaniuk Jul 20 '23 at 20:40