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

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Answer 1 · 2016-11-08T16:09:19

$x = - \frac{27}{40}$ $x=-27/40$

$x < 0$ $x<0$ to make ${log}_{3} (- 5 x)$ $log_3(-5x)$ real.

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

${log}_{3} (- 40 x) = 3$ $log_3(-40x)=3$ .

Inversely,

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

$x = - \frac{27}{40}$ $x=-27/40$ .

How do you solve log_3 (8)+log_3(-5x)=3log3(8)+log3(−5x)=3log_3 (8)+log_3(-5x)=3?