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

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Answer 1 · 2015-10-17T23:51:11

$1 + \sqrt{1 + e} \approx 2.9283$ $1+sqrt(1+e)~~2.9283$

Use the properties of logarithms to rewrite the expression.

$ln (x (x - 2)) = 1$ $ln(x(x-2))=1$

Rewrite both sides in terms of the base $e$ $e$

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

From the properties of logarithms this becomes

$x (x - 2) = e$ $x(x-2)=e$

Distribute $x$ $x$ on left hand side

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

Subtract $e$ $e$ form both sides

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

$\frac{- (- 2) \pm \sqrt{{(- 2)}^{2} - 4 (1) (- e)}}{2} (1)$ $(-(-2)+-sqrt((-2)^2-4(1)(-e)))/2(1)$

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

$\frac{2 \pm \sqrt{4 (1 + e)}}{2}$ $(2+-sqrt(4(1+e)))/2$

$\frac{2}{2} \pm \frac{2 \sqrt{1 + e}}{2}$ $2/2+-(2sqrt(1+e))/2$

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

We are only interested in $1 + \sqrt{1 + e} \approx 2.9283$ $1+sqrt(1+e)~~2.9283$

The other root is negative and not in the domain of the logarithmic function.

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