# How do you condense 3log_2 b - log_2 4 - 2log_2 C?

Jun 14, 2016

$3 {\log}_{2} b - {\log}_{2} \left(4\right) - 2 {\log}_{2} c = {\log}_{2} \left({b}^{3} / \left(4 {c}^{2}\right)\right)$

#### Explanation:

Using $p {\log}_{a} m = {\log}_{a} {m}^{p}$ and $\log a + \log b - \log c = \log \left(\frac{a b}{c}\right)$

$3 {\log}_{2} b - {\log}_{2} \left(4\right) - 2 {\log}_{2} c$

= $\log - 2 \left({b}^{3}\right) - {\log}_{2} \left(4\right) - {\log}_{2} \left({c}^{2}\right)$

= ${\log}_{2} \left({b}^{3} / \left(4 {c}^{2}\right)\right)$

Jun 14, 2016

${\log}_{2} \left({b}^{3} / \left(4 {C}^{2}\right)\right) .$

#### Explanation:

The Rule (1) $m {\log}_{2} n = {\log}_{2} {n}^{m} .$
Rule (2) ${\log}_{2} p - {\log}_{2} q = {\log}_{2} \left(\frac{p}{q}\right) .$

Thus, by Rule (1), $3 {\log}_{2} b = {\log}_{2} {b}^{3} , - 2 {\log}_{2} C = - {\log}_{2} {C}^{2.}$
Combining these by Rule (2) we get, $3 {\log}_{2} b - 2 {\log}_{2} C = {\log}_{2} \left\{\frac{{b}^{3}}{{C}^{2}}\right\} .$

Finally, we get the simplification $= {\log}_{2} \left\{\frac{{b}^{3}}{{C}^{2}}\right\} - {\log}_{2} \left(4\right) = {\log}_{2} \left\{\frac{{b}^{3} / {C}^{2}}{4}\right\} = {\log}_{2} \left({b}^{3} / \left(4 {C}^{2}\right)\right) .$