e^{ln(w^2+1)}=5

Lösung

e^{l n (w^{2} + 1)} = 5

w = 2, w = - 2

Schritte zur Lösung

e^{l n (w^{2} + 1)} = 5

Wende Exponentenregel an

ln (w^{2} + 1) = ln (5)

Löse

ln (w^{2} + 1) = ln (5) : w = 2, w = - 2

w = 2, w = - 2

\frac{3}{3 ^{- k}} = 927

2 \cdot 3^{x} + 6 = 20

so l v e f or x, f^{- 1} (x) = 3 + e^{2 - x}

3 x (1 0^{x}) + 1 0^{x} = 0

2^{3 y - 2} = 16