Question

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

Answer 1

`log_2 ((xy)/4)`

Explanation:

`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)`