How do you solve #log_2(6x+9)-log_2(3)=3#?

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Answer 1 · 2015-12-24T05:33:41

#x=5/2#

Use the following rule: #log_a (b)-log_a (c)=log_a (b/c)#

#log_2(6x+9)-log_2(3)=3#

#=>log_2((6x+9)/3)=3#

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

Exponentiate both sides with base #2# to undo the logarithm.

#2^(log_2(2x+3))=2^3#

#2x+3=8#

#2x=5#

#x=5/2#