Hallo Sarah,
2000·0.5^{0.01·x} = 100·2^{0.01·x} + 100
⇔ 2000·2^{-0.01·x} = 100·2^{0.01·x} + 100
⇔ 2000·(2^{0.01·x}^{-1} = 100·2^{0.01·x}
Substitution z = 20.01·x
2000 * z^{-1} = 100 z + 100 | * z
2000 = 100 z^2 + 100 z | - 2000 | : 100 | ↔
z^2 + z - 20 = 0
pq-Formel → z = 4 oder z = - 5
2^{0.01·x} = 4 [ oder 2^{0.01·x} = - 5 < 0 entfällt ]
2^{0.01·x} = 2^2
0.01·x = 2 | * 100
x = 200
D(200) = S(200) = 500 (!)
Gruß Wolfgang