2(3^{2x-1})=4^{x+1}

Lösung

2 (3^{2 x - 1}) = 4^{x + 1}

x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

Dezimale

x = 2.20951 \dots

Schritte zur Lösung

2 (3^{2 x - 1}) = 4^{x + 1}

Wende Exponentenregel an

ln (2) + (2 x - 1) ln (3) = (x + 1) ln (4)

Löse

ln (2) + (2 x - 1) ln (3) = (x + 1) ln (4) : x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

x = \frac{ln ( 6 )}{2 ( ln ( 3 ) - ln ( 2 ) )}

3^{x + 1} + 3^{x} + 3^{x - 1} = 117

3 (2^{2 x}) = 0.75

5^{x} = 100

6 x = \frac{10}{3} - 6^{- x}

2^{x - 3} = 16