# find the domain of root of a logarithmic function I'm a little confused about this question since o

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 $-\mathrm{\infty }$ to $\mathrm{\infty }$ .I should find the domain of this function: $y=\sqrt{{\mathrm{log}}_{x}\left(10-{x}^{2}\right)}$ . How can I find the interval that makes ${\mathrm{log}}_{x}\left(10-{x}^{2}\right)$ greater than zero?
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Recall for $a>0,a\ne 1,b>0$
${\mathrm{log}}_{a}b=\frac{\mathrm{ln}b}{\mathrm{ln}a}$
Thus we have
$f\left(x\right)={\mathrm{log}}_{x}\left(10-{x}^{2}\right)=\frac{\mathrm{ln}\left(10-{x}^{2}\right)}{\mathrm{ln}x}$
Then $f\left(x\right)\ge 0$ if and only if $10-{x}^{2}\ge 1,x>1$ or $0<10-{x}^{2}\le 1,0. Since the later case cannot happen, then we must have $10-{x}^{2}\ge 1$ and $x>1$, which gives $1

Joel French
First of all, I would suggest to write your $\mathrm{log}$ in a fixed basis
${\mathrm{log}}_{x}\left(10-{x}^{2}\right)=\frac{\mathrm{ln}\left(10-{x}^{2}\right)}{\mathrm{ln}\left(x\right)}.$
$\frac{\mathrm{ln}\left(10-{x}^{2}\right)}{\mathrm{ln}\left(x\right)}>0$
or equivalently:
$\mathrm{ln}\left(x\right)\mathrm{ln}\left(10-{x}^{2}\right)>0.$
Now, you have to find all roots of $10-{x}^{2}$ which gives you the intervals where $\mathrm{ln}\left(10-{x}^{2}\right)>0$ and then you are almost finished.