Solve the equation $ln (x) + ln (x - 3) + ln (2 x - 6) = 0$ $ln(x)+ln(x-3)+ln(2x-6)=0$ ?

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Answer 1 · 2017-07-30T00:14:32

$x \approx 3.3844$ $x ~~ 3.3844$

We have:

$ln (x) + ln (x - 3) + ln (2 x - 6) = 0$ $ln(x)+ln(x-3)+ln(2x-6)=0$

If we look at the graph of:

$y = ln (x) + ln (x - 3) + ln (2 x - 6)$ $y= ln(x)+ln(x-3)+ln(2x-6)$
graph{ln(x)+ln(x-3)+ln(2x-6) [-10, 10, -5, 5]}

Then we appear to have a solution, $α \approx 3.5$ $alpha ~~ 3.5$

First note that for each individual logarithm to exist we require $x$ $x$ to simultaneously satisfy the following inequalities:

$ln(x) in RR => x gt 0$
$ln(x-2) in RR => x-3 gt 0 => x gt 3$
$ln(2x-5) in RR => 2x-6 gt 0 => x gt 3$

Thus $x gt 3$

We can find this solution algebraically:

$ln(x)+ln(x-3)+ln(2x-6)=0$

$:. ln{x(x-3)(2x-6)}=0$
$:. x(x-3)(2x-6) = e^0$
$:. x(x-3)(2x-6) = 1$

And we can multiply out the expression to get:

$x(2x^2-6x-6x+18) = 1$
$:. 2x^3-12x^2+18x = 1$
$:. 2x^3-12x^2+18x - 1 = 0$

As this cubic does not factorise we solve it numerically:

graph{2x^3-12x^2+18x - 1 [-10, 10, -5, 5]}

And we get three real solutions:

$x ~~ 0.0578, 2.5579, 3.3844$

From earlier we established that $x gt 3$ , thus we can eliminate two of these solutions leaving:

$x ~~ 3.3844$

Solve the equation ln(x)+ln(x-3)+ln(2x-6)=0ln(x)+ln(x−3)+ln(2x−6)=0ln(x)+ln(x-3)+ln(2x-6)=0?