17^{x+2}=4

Lösung

1 7^{x + 2} = 4

x = \frac{2 ln ( 2 )}{ln ( 17 )} - 2

Dezimale

x = - 1.51069 \dots

Schritte zur Lösung

1 7^{x + 2} = 4

Wende Exponentenregel an

(x + 2) ln (17) = ln (4)

Löse

(x + 2) ln (17) = ln (4) : x = \frac{2 ln ( 2 )}{ln ( 17 )} - 2

x = \frac{2 ln ( 2 )}{ln ( 17 )} - 2

384 e^{8} = 2^{x}

3^{- 8 x} = 9^{3 - 3 x}

4 \cdot 3^{2 x} + 8 \cdot 3^{x} - 5 = 0

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

e^{3 t} = 4^{9}