How do you solve #log_2(4y-10)>=log_2(y-1)#?

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Answer 1 · 2016-10-18T17:06:39

#log_2(4y - 10) - log_2(y - 1) ≥ 0#

#log_2((4y - 10)/(y - 1)) ≥ 0#

#(4y - 10)/(y - 1) ≥ 2^0#

#4y - 10 ≥ 1(y - 1)#

#3y ≥ 9#

#y ≥ 3#

Hopefully this helps!