ln(1+x^x)=0

Lösung

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

ln (1 + e^{x l n (x)}) = 0

Schritte zur Lösung

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

Wende Exponentenregel an

ln (1 + e^{x l n (x)}) = 0

a^{x - 5} = a

4^{- x} = 2^{x^{2} + 2 x - 8}

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

2 3^{9} - 2 3^{8} = 22 \cdot 2 3^{x}

62 5^{x} = 25