3^{n+1}=2^n

Lösung

3^{n + 1} = 2^{n}

n = - \frac{ln ( 3 )}{ln ( 3 ) - ln ( 2 )}

Dezimale

n = - 2.70951 \dots

Schritte zur Lösung

3^{n + 1} = 2^{n}

Wende Exponentenregel an

(n + 1) ln (3) = n ln (2)

Löse

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

n = - \frac{ln ( 3 )}{ln ( 3 ) - ln ( 2 )}

[(- \frac{2}{3})^{2}]^{x} = 1

1 0^{14} = (1 0^{- 9}) (1 0^{x})

77 = 1 5^{x}

e^{(l n (0.4)) t} = 0.7

so l v e f or n, 128 = 2^{n - 1}