# How do you expand ln ((5sqrty)/x^2)?

Apr 10, 2018

$\textcolor{b l u e}{\frac{1}{2} \ln 5 + \frac{1}{2} \ln y - 2 \ln x}$

#### Explanation:

The law of logarithms state:

${\log}_{c} \left(\frac{a}{b}\right) = {\log}_{c} a - {\log}_{c} b$

${\log}_{c} \left(a b\right) = {\log}_{c} a + {\log}_{c} b$

${\log}_{c} {a}^{b} = b {\log}_{c} a$

Notice we can write:

$5 \sqrt{y} = 5 {y}^{\frac{1}{2}}$

$\therefore$

$\ln \left(\frac{5 \sqrt{y}}{x} ^ 2\right) = \ln \left(\frac{5 {\left(y\right)}^{\frac{1}{2}}}{x} ^ 2\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \ln 5 {\left(y\right)}^{\frac{1}{2}} - \ln {x}^{2}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1}{2} \ln 5 y - 2 \ln x$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \textcolor{b l u e}{\frac{1}{2} \ln 5 + \frac{1}{2} \ln y - 2 \ln x}$