4^{3+x}=7^{1-x}

Lösung

4^{3 + x} = 7^{1 - x}

x = \frac{ln ( 7 ) - 6 ln ( 2 )}{ln ( 28 )}

Dezimale

x = - 0.66411 \dots

Schritte zur Lösung

4^{3 + x} = 7^{1 - x}

Wende Exponentenregel an

(3 + x) ln (4) = (1 - x) ln (7)

Löse

(3 + x) ln (4) = (1 - x) ln (7) : x = \frac{ln ( 7 ) - 6 ln ( 2 )}{ln ( 28 )}

x = \frac{ln ( 7 ) - 6 ln ( 2 )}{ln ( 28 )}

4^{3 - x} = 4

\frac{1}{2} e^{x - 1} = 15

3^{x + 2} = 27

so l v e f or n, y^{n} + y = 0

4^{x} = 8^{x - 1}