How do you evaluate $log_9 (1/3)+ 3 log_9 3$ ?

Question

2103 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-23T14:30:26

Let's first simplify with the rule $alogx = logx^a$ .

$log_9(1/3) + log_9(3^3)$

= $log_9(1/3) + log_9(27)$

Since we're in the same base, we can simplify using the rule $log_an + log_am = log_a(n xx m)$

= $log_9(1/3 xx 27)$

= $log_9(9)$

We can now evaluate using the rule $log_aa = 1$

= $1$

Hopefully this helps!

How do you evaluate log_9 (1/3)+ 3 log_9 3log_9 (1/3)+ 3 log_9 3?