How do you solve $log_3 (8)+log_3(-5x)=3$ ?

Question

1626 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-08T16:09:19

$x=-27/40$

$x<0$ to make $log_3(-5x)$ real.

Use log (ab) = log a + log b.

$log_3(-40x)=3$ .

Inversely,

$-40x=3^3=27$ , and so,

$x=-27/40$ .

How do you solve log_3 (8)+log_3(-5x)=3log_3 (8)+log_3(-5x)=3?