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

1 Answer
May 14, 2016

Answer:

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

Explanation:

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#