Hallo pusteblume,
> 4*e-x + 3ex - 8 = 0
die Substitution wäre z = ex also 1/z = e-x
4/z + 3z - 8 = 0 | * z und dann links "ordnen"
3z2 - 8z + 4 = 0
ax2 + bx + c = 0
abc-Formel: a = 3 , b = - 8 , c = 4
z1,2 = ( - b ± \(\sqrt[]{b^2-4ac}\) ) / (2a)
Einsetzen ergibt:
z1,2 = ( -8 ± √(64 - 48)) / 6 = (8 ± √16 ) / 6 = ( 8 ± 4 ) / 6
z1 = 2 ; z2 = 2/3
Rücksubstitution:
ex = 2 → x1 = ln(2)
ex = 2/3 → x2 = ln(2/3)
Gruß Wolfgang