find the domain of root of a logarithmic function I'm a little confused about this question since output of a logarithmic function varies from $ -\infty $ to $\infty$ .I should find the domain of this function:  $ y=\sqrt{\log_x(10-x^2)} $ . How can I find the interval that makes $\log_x(10-x^2)$ greater than zero?

You first must have $x>1$ because $\log_x(a)=\dfrac{\log(a)}{\log(x)} \implies$ $\log(x)\neq 0$ so $x \neq 1$ which means that $x>1$. Furthermore, the reason it must be greater than $1$ and not simply "not equal to" is to have a positive function for the square root. Notice that one may argue that both $\log(a)$ and $\log(x)$ can be negative and still have a positive function. However, for this example, such cannot happen because for $0<x<1$ the function  $ \ \log(10-x^2)>0$.
Another added constraint is that $10-x^2\geq 1$ in  order to have a a positive square root. Solving for $x$, gives $x\leq 3$ or $x\geq-3$. 
Hence, combining both we have the domain for $x$ as $1< x \leq 3$
