How do you solve #log_12x=1/2log_12 9+1/3log_12 27#?

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Answer 1 · 2017-02-18T12:34:11

#x=9#

#log_12x=1/2log_12 9+1/3log_12 27#

= #log_12 (9^(1/2)xx27^(1/3))#

Hence #x=9^(1/2)xx27^(1/3)#

= #(3^2)^(1/2)xx(3^3)^(1/3)#

= #3^((2xx1/2))xx3^((3xx1/3))#

= #3^1xx3^1#

= #3xx3=9#