Answer
(a) $\ln \frac{\sqrt{x}}{(x^2+1)^2}$
(b) $\ln \frac{1}{x^2-9}$
Work Step by Step
Part (a)
$\frac{1}{2}\ln x-2\ln(x^2+1)$ (Use the property $\ln a^m=m\ln a$)
$=\ln x^{1/2}-\ln(x^2+1)^2$
$=\ln\sqrt{x}-\ln(x^2+1)^2$ (Use the property $\ln \frac{a}{b}=\ln a-\ln b$)
$=\ln \frac{\sqrt{x}}{(x^2+1)^2}$
Part (b)
$\ln (x-3)+\ln (x+3)-2\ln (x^2-9)$
$=\ln (x-3)+\ln (x+3)-2\ln (x^2-9)$ (Use the property $\ln (ab)=\ln a+\ln b$)
$=\ln ((x-3)(x+3))-2\ln (x^2-9)$
$=\ln (x^2-9)-2\ln (x^2-9)$
$=-\ln (x^2-9)$ (Use the property $\ln a^m=m\ln a$)
$=\ln (x^2-9)^{-1}$
$=\ln \frac{1}{x^2-9}$
