ln(x)+ln(x+6)=2in^9

Lösung

ln (x) + ln (x + 6) = 2 i n^{9}

x = - 3 + e^{2 i n^{9}} + 9, x = - e^{2 i n^{9}} + 9 - 3

Schritte zur Lösung

ln (x) + ln (x + 6) = 2 i n^{9}

Wende die log Regel an

x (x + 6) = e^{2 i n^{9}}

Löse

x (x + 6) = e^{2 i n^{9}} : x = - 3 + e^{2 i n^{9}} + 9, x = - e^{2 i n^{9}} + 9 - 3

x = - 3 + e^{2 i n^{9}} + 9, x = - e^{2 i n^{9}} + 9 - 3

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

so l v e f or x, ln^{3} (x) = t^{2}

1 = ln (c)

so l v e f or y, 0.45 = ln (y)

lo g_{2} (2 x) + 2 lo g_{2} (8) = 2