Question

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

Answer 1

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

Explanation:

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`