Answer
(a) $\ln x+\frac{1}{2}\ln (x+1)$
(b) $\frac{1}{2}\log_2(x^2+1)-\frac{1}{2}\log_2(x-1)$
Work Step by Step
Part (a)
$\ln x\sqrt{x+1}=\ln x+\ln \sqrt{x+1}$ (By the property $\ln (ab)=\ln(a)+\ln(b)$)
$=\ln x+\ln (x+1)^{1/2}$
$=\ln x+\frac{1}{2}\ln (x+1)$ (By the property $\ln (a^m)=m\ln (a)$)
Part (b)
$\log_2\sqrt{\frac{x^2+1}{x-1}}=\log_2\left(\frac{x^2+1}{x-1}\right)^{1/2}$
$=\frac{1}{2}\log_2\left(\frac{x^2+1}{x-1}\right)$ (By the property $\log_c (a^m)=m\log_c a$)
$=\frac{1}{2}(\log_2(x^2+1)-\log_2(x-1))$ (By the property $\log_c\frac{a}{b}=\log_c a-\log_c b$)
$=\frac{1}{2}\log_2(x^2+1)-\frac{1}{2}\log_2(x-1)$
