16^{-n}=(1/4)^{3n+1}

Lösung

1 6^{- n} = (\frac{1}{4})^{3 n + 1}

n = - 1

Schritte zur Lösung

1 6^{- n} = (\frac{1}{4})^{3 n + 1}

Wende Exponentenregel an

- 2 n = - (3 n + 1)

Löse

- 2 n = - (3 n + 1) : n = - 1

n = - 1

\frac{1}{5 ^{- 3 x}} = 2 5^{x - 1}

7^{g} = 14

(\frac{5}{7})^{x} = 1

6^{x + 3} = 17

1 0^{x - 3} = 2^{10 + x}