# How do you expand ln (sqrt((3^-2)(5^3)(2^-2)))?

Jan 8, 2016

$\ln \left(\sqrt{{3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2}\right) = \frac{1}{2} \left(3 \ln \left(5\right) - 2 \ln \left(2\right) - 3 \ln \left(3\right)\right)$

#### Explanation:

Given that $\ln \left(x\right) = {\log}_{e} x$
and are satisfied:
$x > 0 , e > 0 , e \ne 1$

We can apply the logarithmic properties:

$\ln \left(a \cdot b\right) = \ln \left(a\right) + \ln \left(b\right)$
$\ln \left(\frac{a}{b}\right) = \ln \left(a\right) - \ln \left(b\right)$
$\ln \left({a}^{b}\right) = b \ln \left(a\right)$

and remembering that:

${a}^{\frac{m}{n}} = \sqrt[n]{{a}^{m}}$
${a}^{-} m = \frac{1}{a} ^ m$

then:

$\ln \left(\sqrt{{3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2}\right) = \ln \left({\left({3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2\right)}^{\frac{1}{2}}\right) =$
$= \frac{1}{2} \ln \left({3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2\right) = \frac{1}{2} \left(\ln \left({3}^{-} 3\right) + \ln \left({5}^{3}\right) + \ln \left({2}^{-} 2\right)\right) =$
$= \frac{1}{2} \left(- 3 \ln \left(3\right) + 3 \ln \left(5\right) - 2 \ln \left(2\right)\right)$

Alternitevely:

$\ln \left({\left({3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2\right)}^{\frac{1}{2}}\right) = \frac{1}{2} \ln \left({3}^{-} 3 \cdot {5}^{3} \cdot {2}^{-} 2\right) =$
$= \frac{1}{2} \ln \left({5}^{3} / \left({3}^{3} \cdot {2}^{2}\right)\right) =$
$= \frac{1}{2} \left(\ln \left({5}^{3}\right) - \ln \left({3}^{3}\right) - \ln \left({2}^{2}\right)\right) =$
$= \frac{1}{2} \left(3 \cdot \ln \left(5\right) - 3 \ln \left(3\right) - 2 \ln \left(2\right)\right)$