How do you expand #ln ((sqrt(a)/bc))#?

1 Answer
Oct 17, 2015

Answer:

#1/2 log(a) + log(c) - log(b)#.

Explanation:

You could write the argument of the logarithm as #sqrt(a) * c * 1/b#. This is useful because the logarithm of a product is the sum of the logarithms, so

#log(sqrt(a) * c * 1/b) = log(sqrt(a)) + log(c) + log(1/b)#

Now we use the rule which states that #log(a^b)=blog(a)#. Writing #sqrt(a)# as #a^{1/2}#, and #1/b# as #b^{-1}#, we get

#log(sqrt(a))=1/2log(a)#, and #log(1/b)=-log(b)#.