Given log2=.3010 and log3+.4771, how do you evaluate #log(3/2)#?

Question

2881 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-04T02:30:26

#log(3/2) = log(3)-log(2) = 0.4771-0.3010=0.1761#

Let's review a few important properties of logarithms, easily derived from its definition: #log_b(p)# is a number #a# such that #b^a = p#.

(1) #log_b(1/q)=-log_b(q)#
because #b^(-log_b(q)) = 1/b^(log_b(q))=1/q#

(2) #log_b(p*q)=log_b(p)+log_b(q)#
because #b^(log_b(p)+log_b(q))=b^(log_b(p))*b^(log_b(q))=p*q#

(3) #log_b(p/q)=log_b(p)-log_b(q)#
because #b^(log_b(p)-log_b(q)) = b^(log_b(p))/b^(log_b(q))=p/q#

(4) #log_a(p)=log_b(p)*log_a(b)#
because #a^(log_b(p)*log_a(b))=(a^(log_a(b)))^(log_b(p))=#
#=b^(log_b(p)) = p#

You can get more details about logarithms at Unizor by following the menu options Algebra - Logarithmic Functions.

Using the property (3) above,
#log(3/2) = log(3)-log(2) = 0.4771-0.3010=0.1761#