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?
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$\begingroup$Recall for $a>0, a\neq 1,b>0$:
$$\log_ab=\frac{\ln b}{\ln a}$$
Thus we have
$$f(x)=\log_x(10-x^2)=\frac{\ln(10-x^2)}{\ln x}$$
Then $f(x)\geq 0$ if and only if $10-x^2\geq 1,x>1$ or $0<10-x^2\leq 1,0<x<1$. Since the later case cannot happen, then we must have $10-x^2\geq 1$ and $x>1$, which gives $1<x\leq3$
$\endgroup$ $\begingroup$First of all, I would suggest to write your $\log$ in a fixed basis $$\log_x(10-x^2) = \frac{\ln(10-x^2)}{\ln(x)}.$$
This changes your question to $$\frac{\ln(10-x^2)}{\ln(x)} > 0$$ or equivalently: $$\ln(x)\ln(10-x^2) > 0.$$
Now, you have to find all roots of $10-x^2$ which gives you the intervals where $\ln(10-x^2) > 0$ and then you are almost finished.
$\endgroup$ $\begingroup$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 $1< x \leq 3$
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