5^{x+1}=4^{2x-3}

Lösung

5^{x + 1} = 4^{2 x - 3}

x = \frac{ln ( 320 )}{4 ln ( 2 ) - ln ( 5 )}

Dezimale

x = 4.95922 \dots

Schritte zur Lösung

5^{x + 1} = 4^{2 x - 3}

Wende Exponentenregel an

(x + 1) ln (5) = (2 x - 3) ln (4)

Löse

(x + 1) ln (5) = (2 x - 3) ln (4) : x = \frac{ln ( 320 )}{4 ln ( 2 ) - ln ( 5 )}

x = \frac{ln ( 320 )}{4 ln ( 2 ) - ln ( 5 )}

8^{2 x + 1} = 105

so l v e f or x, e^{\frac{x}{2}} = 4 l n x

2^{- b} = 1

5^{5 - x} = \frac{1}{25}

so l v e f or x, 312 5^{x + 4} = 2 5^{2 x}