## Precalculus (6th Edition) Blitzer

$log_b(\frac{\sqrt x}{x^2+1})$
Using logarithmic properties, we have $\frac{1}{2}log_b(x)-log_b(x^2+1)=log_b(\sqrt x)-log_b(x^2+1)=log_b(\frac{\sqrt x}{x^2+1})$