6(3^{2x})=2^{x+1}

Lösung

6 (3^{2 x}) = 2^{x + 1}

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

Dezimale

x = - 0.73042 \dots

Schritte zur Lösung

6 (3^{2 x}) = 2^{x + 1}

Wende Exponentenregel an

ln (6) + 2 x ln (3) = (x + 1) ln (2)

Löse

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

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

ln (3) = e^{x}

(2^{x} - 3) (2^{x} - 4) = 0

9^{- x + 15} = 2 7^{x}

312 5^{x} = e

2 5^{2 x - 3} = 12 5^{x + 4}