How do you simplify #log_2 14 - log_2 7#?

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Answer 1 · 2016-05-14T14:21:16

#log_2(14) - log_2(7) = 1#

Using the log rule #log_x(a) – log_x(b) = log_x(a/b)#

Rewrite the equation as:

#log_2(14/7)# = #log_2(2)#

Use the log rule:

#log_x(x)# = 1

Therefore #log_2(2)# = 1

So #log_2(14) - log_2(7) = 1#