3^{2x}=9^{3x+1}

Lösung

3^{2 x} = 9^{3 x + 1}

x = - \frac{1}{2}

Dezimale

x = - 0.5

Schritte zur Lösung

3^{2 x} = 9^{3 x + 1}

Wende Exponentenregel an

2 x = 2 (3 x + 1)

Löse

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

x = - \frac{1}{2}

lo g_{5} (7^{3 x}) = 10

4^{- \frac{1}{x}} = \frac{1}{2}

5. 5^{r} = 1.6

2^{2 x} - 29 (2^{x}) - 96 = 0

9^{1 - 2 x} = \frac{1}{2 7 ^{x - 4}}