How do you condense $log_2xy-log_x x^2$ ?

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Answer 1 · 2018-08-04T15:57:30

$log_2 ((xy)/4)$

$log_2xy-log_x x^2$

$=log_2x+log_2y-log_2x^2/log_2x$

$=log_2x+log_2y- (2log_2x)/log_2x$

$=log_2x+log_2y-2$

$=log_2xy-log_2 2^2$

$=log_2 ((xy)/4)$

How do you condense log_2xy-log_x x^2log_2xy-log_x x^2?