Find the domain of the function $$g(x)=\log_3(x^2-1)$$
This is what I got so far:
$$\{ x\mid x^2-1>0\} =$$ $$\{ x\mid x^2>1\} =$$ $$\{ x\mid x>\sqrt { 1 } \}= $$
I don't know where to go from here to arrive at the correct answer... I would like a nudge in the right direction, if possible.
I would also like to know why it is that with any logarithmic function, $\log x$ is defined when $x>0$. It isn't really "clicking" in my head for me.
On
$x^2>1 \iff (x<-1$ or $x>1$) Alternatively, look at the graph of $x^2-1$ and see where it is above the $x-$axis
To answer your other question, logarithms are exponents. $\log_3(x)$ is the exponent $y$ which makes $3^y=x$. Now if $x$ were negative, say $x=-1$, then what is the exponent $y$ such that $3^y=-1.$ Of course there is none. So $\log_3(-1)$ doesnt make sense!
What is in the argument of $\log_3$? It is $x^2 - 1$, right? So you need the values of $x$ such that: $$x^2 - 1 > 0$$ We have: $$x^2 - 1 > 0 \implies x^2 > 1 \implies |x| > 1$$ You got it wrong there because $\sqrt{x^2} = |x|$, so you let escape a few possibilities. Now, drawing the real line, it is easy to see that $|x| > 1$ happens if and only if $x > 1$, or $x < -1$. Ok?
We have that $\log_a b = x \iff a^x = b$, by definition. If $a > 0$ and $b < 0$, you get something positive equaling something negative.