How do you condense #log_4(xy)^3-log_4(xy)#?

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Answer 1 · 2018-07-06T15:22:30

#color(cyan )(=> log (xy)^2#

#log_4 (xy)^3 - log_4 (xy)#

#log x^n = blog x#

#=> 3 log_4 (xy) - log_4 (xy)#

#=> 2 log (xy)#

#=> log (xy)^2#

Answer 2 · 2018-07-06T15:29:45

#2log(xy)=log(xy)^2#

Given: #log_4(xy)^3-log_4(xy)#

Use the fact that #log_bm^n=nlog_bm#.

#=3log_4(xy)-log_4(xy)#

Let #u=log_4(xy)#.

#=3u-u#

#=2u#

Substitute back #u=log_4(xy)# to get:

#=2log(xy)#

Alternatively, we can bring the exponent back, and it will give us:

#=log(xy)^2#