2^{2x+1}=5

Lösung

2^{2 x + 1} = 5

x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

Dezimale

x = 0.66096 \dots

Schritte zur Lösung

2^{2 x + 1} = 5

Wende Exponentenregel an

(2 x + 1) ln (2) = ln (5)

Löse

(2 x + 1) ln (2) = ln (5) : x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

x = \frac{ln ( 5 ) - ln ( 2 )}{2 ln ( 2 )}

3 \cdot 9^{x} + 8 \cdot 3^{x} - 3 = 0

12 5^{x} = \frac{1}{5}

e^{x} - 6 x = 0

3 e^{x} - 4 x e^{x} = 0

3 \cdot 1 0^{- 9} = 1 0^{- x}