How do you solve #log_3(x+16) - log_3(x) = log_3(2)#?

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Answer 1 · 2016-06-07T02:36:17

#x=16#

First, note that they are all in #log_3#, so we simplify the left-hand side to get

#log_3((x+16)/x) = log_3 2#
#=> (x+16)/x = 2#, #2x=x+16#, #x=16#

Answer 2 · 2016-06-07T02:37:55

#16=x#

The log laws suggest that when to log's with the same base subtract from one another, it can be written as

#log_a(x)-log_a(y)=log_a(x/y)#

Therefore, our equation can be simplified into

#log_3((x+16)/x)=log_3(2)#

Therefore #(x+16)/x=2#

#x+16=2x#

#16=x#