How do you expand this logarithm?

${\log}_{3} \left({z}^{4} \sqrt{x}\right)$

Nov 21, 2016

$4 {\log}_{3} \left(z\right) + \frac{1}{2} {\log}_{3} \left(x\right)$

Explanation:

General Rules:
$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} {a}^{c} = c \cdot {\log}_{b} a$

$\textcolor{w h i t e}{\text{XXX}} {\log}_{b} \left(a \cdot c\right) = {\log}_{b} + {\log}_{b} c$

Nov 21, 2016

$4 {\log}_{3} \left(z\right) + \frac{1}{2} {\log}_{3} \left(x\right)$.

Explanation:

By using the rule of logarithms where $\log \left(a \cdot b\right) = \log a + \log b$, we first get

${\log}_{3} \left({z}^{4} \sqrt{x}\right)$
$= {\log}_{3} \left({z}^{4}\right) + {\log}_{3} \left(\sqrt{x}\right)$
$= {\log}_{3} \left({z}^{4}\right) + {\log}_{3} \left({x}^{1 / 2}\right)$

Another rule of logarithms is $\log \left({a}^{b}\right) = b \log \left(a\right)$. We now use this to get

$= 4 {\log}_{3} \left(z\right) + \frac{1}{2} {\log}_{3} \left(x\right)$

Unless you have been asked to rewrite the "base 3" logarithms in "base 10" form, this is as much expansion as we can do.

Hope this helps!