How do you solve #\log _ { 2} x + \log _ { 2} ( x - 2) > 3#?

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Answer 1 · 2018-01-26T09:58:02

#x>4#

Thedomain of #log_2x+log_2(x-2)>3# is #x>2#

Now #log_2x+log_2(x-2)>3#

#=>log_2(x(x-2))>3#

or #x^2-2x>2^3#

or #x^2-2x-8>0#

or #(x-4)(x+2)>0#

This means that #(x-4)(x+2)# is positive and this happens when

(1) both #x-4>0# and #x+2>0# i.e. #x>4# and #x>-2# i.e. when #x>4#

or (2) both #x-4<0# and #x+2<0# i.e. #x<4# and #x<-2# i.e. when #x<-2#.

But as #x<-2# is not in domain, answe is #x>4#