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

Lösung

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

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

Dezimale

x = 3.81884 \dots

Schritte zur Lösung

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

Wende Exponentenregel an

2 x ln (2) = (x + 1) ln (3)

Löse

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

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

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

5^{2 x - 4} = 125

so l v e f or x, t (x) = 2. 3^{x - 1.5}

4^{x - 2} = 3

1 8^{2 x - 3} = 2 0^{x + 1}