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 $x=9^(1/2)xx27^(1/3)$

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

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

= $3^1xx3^1$

= $3xx3=9$

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