How do you condense #Logx+log2#?

1 Answer
Mar 20, 2016

Answer:

#log(x)+log(2)=log(2x)#

Explanation:

In general, #log_a(x)+log_a(y) = log_a(xy)# (see below for explanation).

Applying this, we get #log(x)+log(2)=log(2x)#


To see why the above property holds, note that #log_a(x)# is the value such that #a^(log_a(x)) = x#. Then, as we have

#a^(log_a(x)+log_a(y)) = a^(log_a(x))a^(log_a(y)) = xy#

we see that #log_a(x)+log_a(y)# is the value which, when taking #a# to that power, results in #xy#. Thus we have

#log_a(x)+log_a(y) = log_a(xy)#