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

1 Answer
Mar 23, 2016

Answer:

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

Explanation:

#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!