How do you solve #3log_3 4a=3#?

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Answer 1 · 2016-06-09T01:05:22

#log_3(4a)^3 = 3#

Use the rule #alogn = logn^a#

#log_3(64a^3) = 3#

Recall that if #log_a(n) = x#, then #a^x = n#

#64a^3 = 27#

#a^3 = 27/64#

#a = root(3)(27/64)#

#a = 3/4#

Hopefully this helps!