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

Lösung

ln (x^{2} + 2) - ln (x - 1) = 2

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

Dezimale

x = 5.75862 \dots, x = 1.63043 \dots

Schritte zur Lösung

ln (x^{2} + 2) - ln (x - 1) = 2

Füge

ln (x - 1)

zu beiden Seiten hinzu

ln (x^{2} + 2) - ln (x - 1) + ln (x - 1) = 2 + ln (x - 1)

Vereinfache

ln (x^{2} + 2) = 2 + ln (x - 1)

Wende die log Regel an

x^{2} + 2 = e^{2} (x - 1)

Löse

x^{2} + 2 = e^{2} (x - 1) : x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

Überprüfe die Lösungen:

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}

Wahr

, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

Wahr

Die Lösungen sind

x = \frac{e ^{2} + e ^{4} - 4 ( 2 + e ^{2} )}{2}, x = \frac{e ^{2} - e ^{4} - 4 ( 2 + e ^{2} )}{2}

15 = 6 lo g_{4} (8 x)

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

lo g_{3} (x + 17) - 2 = lo g_{3} (2 x)

4 = \frac{ln ( 2 x )}{ln ( 4 x )}

2 lo g_{2} (x) = 4