# How do you simplify ln(8x)^(1/2)+ln4x^2-ln(16x)^(1/2)?

May 17, 2018

$\ln {\left(8 x\right)}^{\frac{1}{2}} + \ln 4 {x}^{2} - \ln {\left(16 x\right)}^{\frac{1}{2}} = 2 \ln x + \frac{3}{2} \ln 2$

#### Explanation:

We will use identities $\ln a + \ln b = \ln \left(a \cdot b\right)$, $\ln a - \ln c = \ln \left(\frac{a}{c}\right)$ and $\ln {a}^{n} = n \ln a$.

Hence $\ln {\left(8 x\right)}^{\frac{1}{2}} + \ln 4 {x}^{2} - \ln {\left(16 x\right)}^{\frac{1}{2}}$

= $\ln \left(\frac{{\left(8 x\right)}^{\frac{1}{2}} \cdot 4 {x}^{2}}{16 x} ^ \left(\frac{1}{2}\right)\right)$

= $\ln \left(\frac{{8}^{\frac{1}{2}} \textcolor{red}{{x}^{\frac{1}{2}}} \cdot 4 {x}^{2}}{{16}^{\frac{1}{2}} \textcolor{red}{{x}^{\frac{1}{2}}}}\right)$

= $\ln \left(\frac{{8}^{\frac{1}{2}} \cdot 4 {x}^{2}}{16} ^ \left(\frac{1}{2}\right)\right)$

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

= $\ln {2}^{2} + \ln {x}^{2} - \ln {2}^{\frac{1}{2}}$

= $2 \ln 2 + 2 \ln x - \frac{1}{2} \ln 2$

= $2 \ln x + \frac{3}{2} \ln 2$

May 17, 2018

color(indigo)(=> (3/2) ln 2 + 2 ln x

#### Explanation:

$\log m + \log n = \log \left(m n\right)$

$\log m - \log n = \log \left(\frac{m}{n}\right)$

Given : $\ln {\left(8 x\right)}^{\frac{1}{2}} + \ln \left(4 {x}^{2}\right) - \ln {\left(16 x\right)}^{\frac{1}{2}}$

$\implies \ln \left({\left(8 x\right)}^{\frac{1}{2}} \cdot \left(4 {x}^{2}\right)\right) - \ln {\left(16 x\right)}^{\frac{1}{2}}$

$\implies \ln \left(\frac{2 \sqrt{2} \cdot 4 \cdot {x}^{\frac{1}{2}} \cdot {x}^{2}}{16 x} ^ \left(\frac{1}{2}\right)\right)$

$\implies \ln \left(\frac{{\cancel{8}}^{2} \sqrt{2} \cdot \cancel{{\left(x\right)}^{\frac{1}{2}}} \cdot {x}^{2}}{\cancel{4} \cdot \cancel{{\left(x\right)}^{\frac{1}{2}}}}\right)$

$\implies \ln \left(2 \sqrt{2} {x}^{2}\right)$

$\implies \ln \left({2}^{\frac{3}{2}}\right) + \ln {x}^{2}$

color(indigo)(=> (3/2) ln 2 + 2 ln x