How do you solve $ln x + ln (x - 2) = 1$ $ln x + ln(x-2) = 1$ ?

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How do you solve $ln x + ln (x - 2) = 1$ $ln x + ln(x-2) = 1$ ?

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Answer 1 · 2015-09-10T16:28:21

Use exponentiation to produce a quadratic equation to solve using the quadratic formula.

Discard the root of this quadratic that results in $x < 0$ $x < 0$ , leaving the solution:

$x = 1 + \sqrt{1 + e}$ $x = 1 + sqrt(1+e)$

Explanation:

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

So $x^{2} - 2 x - e = 0$ $x^2 - 2x - e = 0$

Then using the quadratic formula, with $a = 1$ $a = 1$ , $b = - 2$ $b = -2$ and $c = e$ $c = e$

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{2 \pm \sqrt{2^{2} - (4 \times 1 \times - e)}}{2 \cdot 1}$ $x = (-b +-sqrt(b^2-4ac))/(2a) = (2 +-sqrt(2^2-(4xx1xx-e)))/(2*1)$

$= \frac{2 \pm \sqrt{4 + 4 e}}{2}$ $=(2+-sqrt(4+4e))/2$

$= 1 \pm \sqrt{1 + e}$ $=1 +-sqrt(1+e)$

Now $\sqrt{1 + e} > 1$ $sqrt(1+e) > 1$ , so $1 - \sqrt{1 + e} < 0$ $1 - sqrt(1+e) < 0$ and $ln (1 - \sqrt{1 + e})$ $ln(1 - sqrt(1+e))$ is not defined.

So the only valid solution is $x = 1 + \sqrt{1 + e}$ $x = 1 + sqrt(1+e)$

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How do you solve $ln x + ln (x - 2) = 1$ $ln x + ln(x-2) = 1$ ?

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