Answer
$\ln (\frac{8x^2}{\sqrt{x+4}})$
Work Step by Step
$Combine$ $the$ $expression$:
$3$$\ln 2$ + $2$$\ln x$ - $\frac{1}{2}$$\ln (x+4)$
Apply the Third Law of Logarithms for $3$$\ln 2$, $2$$\ln x$, and $\frac{1}{2}$$\ln (x+4)$
$3$$\ln 2$ = $\ln 2^3$ = $\ln 8$
$2$$\ln x$ = $\ln x^2$
$\frac{1}{2}$$\ln (x+4)$ = $\ln (x+4)^\frac{1}{2}$
$\ln 8$ + $\ln x^2$ - $\ln (x+4)^\frac{1}{2}$
$\ln 8$ + $\ln x^2$ - $\ln \sqrt{x+4}$
Apply the First Law of Logarithms for $\ln 8$ + $\ln x^2$
$\ln 8$ + $\ln x^2$ = $\ln (8\times x^2)$
$\ln 8x^2$ - $\ln \sqrt{x+4}$
Apply the Second Law of Logarithms
$\ln 8x^2$ - $\ln \sqrt{x+4}$ = $\ln (\frac{8x^2}{\sqrt{x+4}})$
