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

Question

1529 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-25T14:02:03

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

We can simplify the left hand side by using the logarithmic rule: $ln (a) + ln (b) = ln (a b)$ $ln(a)+ln(b)=ln(ab)$

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

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

To undo the natural logarithm, exponentiate both sides with base $e$ $e$ .

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

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

Move the $e$ $e$ to the left side and solve with the quadratic equation.

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

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

Factor a $4$ $4$ from inside the square root, which can be pulled out as a $2$ $2$ .

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

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

Notice that the negative root has been taken away, since for $ln (a), a > 0$ $ln(a),a>0$ (if we are forgoing complex roots).

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