log_{10}(3x)<log_{10}(7x-8)

Lösung

lo g_{10} (3 x) < lo g_{10} (7 x - 8)

x > 2

Intervall-Notation

(2, \infty)

Schritte zur Lösung

lo g_{10} (3 x) < lo g_{10} (7 x - 8)

Wende die log Regel an

3 x < 7 x - 8 and 3 x > 0

3 x < 7 x - 8 and 3 x > 0 : x > 2

x > 2

s im pl i f y \frac{1}{4} + \frac{1}{20}

6 q^{2} - 3 q = 5 q^{2} - 8 q + 24

m^{2} + 6 m + 9 = 0

x + 1 + 4 \leq 9

3 - (2)^{4}