e^{x+1}=2

Lösung

e^{x + 1} = 2

x = ln (2) - 1

Dezimale

x = - 0.30685 \dots

Schritte zur Lösung

e^{x + 1} = 2

Wende Exponentenregel an

x + 1 = ln (2)

Löse

x + 1 = ln (2) : x = ln (2) - 1

x = ln (2) - 1

ln (\frac{e ^{2 x}}{4}) = 1

4^{x} + 2^{x} = 1

102 4^{x + 1} = 1 6^{5}

so l v e f or x, f^{8 x + 11} = a^{9 x + 4}

e^{x} = 21