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

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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!