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

Jun 4, 2016

$\log \left(\frac{3}{2}\right) = \log \left(3\right) - \log \left(2\right) = 0.4771 - 0.3010 = 0.1761$

Explanation:

Let's review a few important properties of logarithms, easily derived from its definition: ${\log}_{b} \left(p\right)$ is a number $a$ such that ${b}^{a} = p$.

(1) ${\log}_{b} \left(\frac{1}{q}\right) = - {\log}_{b} \left(q\right)$
because ${b}^{- {\log}_{b} \left(q\right)} = \frac{1}{b} ^ \left({\log}_{b} \left(q\right)\right) = \frac{1}{q}$

(2) ${\log}_{b} \left(p \cdot q\right) = {\log}_{b} \left(p\right) + {\log}_{b} \left(q\right)$
because ${b}^{{\log}_{b} \left(p\right) + {\log}_{b} \left(q\right)} = {b}^{{\log}_{b} \left(p\right)} \cdot {b}^{{\log}_{b} \left(q\right)} = p \cdot q$

(3) ${\log}_{b} \left(\frac{p}{q}\right) = {\log}_{b} \left(p\right) - {\log}_{b} \left(q\right)$
because ${b}^{{\log}_{b} \left(p\right) - {\log}_{b} \left(q\right)} = {b}^{{\log}_{b} \left(p\right)} / {b}^{{\log}_{b} \left(q\right)} = \frac{p}{q}$

(4) ${\log}_{a} \left(p\right) = {\log}_{b} \left(p\right) \cdot {\log}_{a} \left(b\right)$
because ${a}^{{\log}_{b} \left(p\right) \cdot {\log}_{a} \left(b\right)} = {\left({a}^{{\log}_{a} \left(b\right)}\right)}^{{\log}_{b} \left(p\right)} =$
$= {b}^{{\log}_{b} \left(p\right)} = p$

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

Using the property (3) above,
$\log \left(\frac{3}{2}\right) = \log \left(3\right) - \log \left(2\right) = 0.4771 - 0.3010 = 0.1761$