16^{2x+1}=3^{3x-4}

Lösung

1 6^{2 x + 1} = 3^{3 x - 4}

x = - \frac{4 ln ( 6 )}{8 ln ( 2 ) - 3 ln ( 3 )}

Dezimale

x = - 3.18628 \dots

Schritte zur Lösung

1 6^{2 x + 1} = 3^{3 x - 4}

Wende Exponentenregel an

(2 x + 1) ln (16) = (3 x - 4) ln (3)

Löse

(2 x + 1) ln (16) = (3 x - 4) ln (3) : x = - \frac{4 ln ( 6 )}{8 ln ( 2 ) - 3 ln ( 3 )}

x = - \frac{4 ln ( 6 )}{8 ln ( 2 ) - 3 ln ( 3 )}

(- 4)^{x} = \frac{1}{16}

1 = 3^{x}

e^{2 x} + e^{x} - 6 = 0

2^{x} + 2^{3 - x} - 9 = 0

so l v e f or k, k (x) = x^{5} e^{4 x}