What is the answer to ln(x)+ln(x+6)=2in^9 ?

The answer to ln(x)+ln(x+6)=2in^9 is x=-3+sqrt(e^{2in^9)+9},x=-sqrt(e^{2in^9)+9}-3

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

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

Solution steps

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

Apply log rules

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

Solve

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)

x, ln^{3} (x) = t^{2}

1 = ln (c)

y, 0.45 = ln (y)

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

What is the answer to ln(x)+ln(x+6)=2in^9 ?
The answer to ln(x)+ln(x+6)=2in^9 is x=-3+sqrt(e^{2in^9)+9},x=-sqrt(e^{2in^9)+9}-3