How do you solve #log_2(x+5)-log_2(x-2)=3#?

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Answer 1 · 2016-07-24T15:08:34

x=3

#log_2(x+5)-log_2(x-2)=3#

combine the log terms as follows

#log_2 ((x+5)/(x-2))=3#

then by definition

#((x+5)/(x-2))=2^3 = 8#

then some algebra

#x+5= 8x-16#

#7x = 21, x=3#