2^{x+1}=5^{x-2}

Lösung

2^{x + 1} = 5^{x - 2}

x = \frac{ln ( 50 )}{ln ( 5 ) - ln ( 2 )}

Dezimale

x = 4.26941 \dots

Schritte zur Lösung

2^{x + 1} = 5^{x - 2}

Wende Exponentenregel an

(x + 1) ln (2) = (x - 2) ln (5)

Löse

(x + 1) ln (2) = (x - 2) ln (5) : x = \frac{ln ( 50 )}{ln ( 5 ) - ln ( 2 )}

x = \frac{ln ( 50 )}{ln ( 5 ) - ln ( 2 )}

\frac{1}{4 ^{x}} = 8

2 e^{x} = 5 e^{x} + 12

so l v e f or t, 5^{t} = 7

12 5^{3 x - 4} = 2 5^{4 x + 2}

5 \cdot 2^{4 x} = 15