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

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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 $x < 0$ , leaving the solution:

$x = 1 + sqrt(1+e)$

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

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

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

$x = (-b +-sqrt(b^2-4ac))/(2a) = (2 +-sqrt(2^2-(4xx1xx-e)))/(2*1)$

$=(2+-sqrt(4+4e))/2$

$=1 +-sqrt(1+e)$

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

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

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