How do you solve #2log_6(4x)=0#?

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Answer 1 · 2016-07-15T21:23:43

#x = 1/4#

By definition if #log_a b= c, " then " a^c = b#

#2log_6 (4x) =0#

#(2log_6 (4x))/2 =0/2#

#log_6 4x = 0" "rArr 6^0 = 4x#

#6^0 = 4x#

#1 = 4x#

#x = 1/4#