What is the domain of this function? (Don't know how to solve it, logarithms...)

Question

Please explain how you solved it, thanks.

$f(x)=\sqrt{\log_x2 - \log_2x}$

Answer 1

First $x$ should be positive and $x\ne1$. Moreoer $$\log_x2-\log_2x=\frac{\log2}{\log x}-\frac{\log x}{\log 2}=\frac{\log^22-\log^2 x}{\log 2\log x}=\frac{(\log2-\log x)(\log2+\log x)}{\log 2\log x}\ge0\\\iff x\in [0,\frac12)\cup(1,2]$$ so the domain of definition is $[0,\frac12)\cup(1,2]$.

Answer 2

Hint

Going, from definition, to natural logarithms, you have $$f(x)=\sqrt{\log_x2 - \log_2x}=\sqrt{\frac{\log (2)}{\log (x)}-\frac{\log (x)}{\log (2)}}$$

I am sure you can take from here.