solvefor x,e^{x^2+x+In(4)}=4

Lösung

löse nach

x, e^{x^{2} + x + I n (4)} = 4

x = \frac{- 1 + 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}, x = \frac{- 1 - 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}

Schritte zur Lösung

e^{x^{2} + x + I n (4)} = 4

Wende Exponentenregel an

x^{2} + x + I n \cdot 4 = ln (4)

Löse

x^{2} + x + I n \cdot 4 = ln (4) : x = \frac{- 1 + 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}, x = \frac{- 1 - 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}

x = \frac{- 1 + 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}, x = \frac{- 1 - 1 - 4 ( 4 n I - 2 ln ( 2 ) )}{2}

150 = 350 - 335 e^{15 x}

so l v e f or x, y^{1} - x e^{x} = 0

2^{x + 3} + 2^{3} = 9

61 = 1 7^{x}

e^{x} = ln (2)