# How do you condense 1/ 7 ln(x + 2)^7 + 1/ 2 ln x − ln(x^2 + 3x + 2)^2?

Sep 14, 2016

$\ln \left(\frac{{x}^{\frac{1}{2}} \left(x + 2\right)}{{x}^{2} + 3 x + 2} ^ 2\right)$

#### Explanation:

Using the $\textcolor{b l u e}{\text{laws of logarithms}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log {x}^{n} \Leftrightarrow n \log x} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\Rightarrow \frac{1}{7} \ln {\left(x + 2\right)}^{7} = \ln {\left(x + 2\right)}^{7 \times \frac{1}{7}} = \ln \left(x + 2\right)$

and $\frac{1}{2} \ln x = \ln {x}^{\frac{1}{2}}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log x + \log y = \log \left(x y\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\log x - \log y = \log \left(\frac{x}{y}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

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