How do you solve # log_2 (x - 2) + 5 = 8 - log_2 4#?

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Answer 1 · 2015-07-18T10:59:41

x=4

#log(ab)=loga+logb#____(1)

#log_a(b)=x =>a^x=b#_______(2)

#log_2(x-2)+5=8-log_2(4)#

#=>log_2(x-2)+log_2(4)=3#

#=>log_2((x-2)4)=3# [since from (1)]

#=>2^3=4(x-2)#[since from (2)]

#=>8=4x-8#

#=>4x=16#

#=>x=4#