How do you solve $ln x + ln (x-3) =0$ ?

Question

8342 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-29T20:51:23

$x = (3 + sqrt(13))/2$

$log_e x+log_e(x-3) = log_e 1$ so
$log_e x(x-3) = log_e 1$

or

$x(x-3)=1 -> x^2-3x-1=0$

with the solutions

$x = (3 pm sqrt(13))/2$ now, cheking the feasibility we choose

$x = (3 + sqrt(13))/2$ because with this value,

$x > 0$ and $x-3 > 0$

How do you solve ln x + ln (x-3) =0ln x + ln (x-3) =0?