$$\frac{g_s-g_e}{g_e-r} \cdot(e^{(g_e-r)\cdot T}-e^{(g_e-r)\cdot S}) = e^ {(g_e-r)\cdot S}$$
soll ergeben:
$$(g_s-g_e)\cdot e^{(g_e-r)\cdot T}= (g_s-r)\cdot e^{(g_e-r)S}$$
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$$\frac{g_s-g_e}{g_e-r} \cdot(e^{(g_e-r)\cdot T}-e^{(g_e-r)\cdot S}) = e^ {(g_e-r)\cdot S} ~~~|\cdot (g_e-r)$$
$$(g_s-g_e)\cdot(e^{(g_e-r)\cdot T}-e^{(g_e-r)\cdot S}) = (g_e-r)e^ {(g_e-r)\cdot S}$$
$$(g_s-g_e)\cdot e^{(g_e-r)\cdot T}-(g_s-g_e)e^{(g_e-r)\cdot S}= (g_e-r)e^ {(g_e-r)\cdot S}~~~~| +(g_s-g_e)e^{(g_e-r)\cdot S}) $$
$$(g_s-g_e)\cdot e^{(g_e-r)\cdot T}=(g_s-g_e)e^{(g_e-r)\cdot S}+(g_e-r)e^ {(g_e-r)\cdot S}~~~~ $$
Und den Rest sieht man jetzt ja schon.
$$(g_s-g_e)\cdot e^{(g_e-r)\cdot T}= (g_s-r)\cdot e^{(g_e-r)S}$$
:-)