2^{x^2+2x+1}=3

Lösung

2^{x^{2} + 2 x + 1} = 3

x = \frac{- ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}, x = - \frac{ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}

Dezimale

x = 0.25895 \dots, x = - 2.25895 \dots

Schritte zur Lösung

2^{x^{2} + 2 x + 1} = 3

Wende Exponentenregel an

(x^{2} + 2 x + 1) ln (2) = ln (3)

Löse

(x^{2} + 2 x + 1) ln (2) = ln (3) : x = \frac{- ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}, x = - \frac{ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}

x = \frac{- ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}, x = - \frac{ln ( 2 ) + ln ( 2 ) ln ( 3 )}{ln ( 2 )}

3^{2 x} - 30 (3^{x}) + 81 = 0

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

8^{3 - x} = 4 \cdot 2^{1 - 2 x}

\frac{1}{49} = 7^{n}

2 5^{x} - 6 \cdot 5^{x} - 16 = 0