<s>[INST] What is the set of values of $a$ ($a \in \mathbb{R}$) for which $\log_a(x^2 + 4) \ge 2$, $\forall x \in \mathbb{R}$? How can I find the set of values of $a$, with $a \in \mathbb{R}$ such that:
$$\log_a(x^2+4) \ge 2, \forall x \in \mathbb{R}$$
How should I approach this?
 [/INST] Note if $0 < a < 1$ then $\log_a M$ is decreasing; If $M > 1$ then $\log_a M < 0$, $\log_a 1 = 0$ and if $0< M < 1$ then $\log_a M > 0$.  And $M < N\iff M> N$.
So $\log_a (x^2 +4) \ge 2 \iff a^{\log_a x^2 +4 } = x^2 + 4 \le a^2 \iff x^2 \le a^2 -4$.  But because $a^2 < 1$  then $a^2 -4 < -3$ and $x^3 < -3$ is impossible.  So $\log_a(x^2 + 4) < 2$ for all $0< a < 1$.
And if $a > 1$ then $\log_a M$ is increasing; If $M > 1$ then $\log_a M > 0$, $\log_a 1 = 0$ and if $0< M < 1$ then $\log_a M < 0$.  And $M < N\iff M< N$.
So $\log_a (x^2 +4) \ge 2 \iff a^{\log_a x^2 +4 } = x^2 + 4 \ge a^2 \iff x^2 \ge a^2 -4$. 
For this be be true for all $x$ then $x^2 \ge 0$ and $x^2 =0$ is possible.  So this is only possible and is always possible if $x^2 \ge 0 \ge a^2 -4$. 
Or in other words if $a^2 \le 4$ or $|a| \le 2$. But as $a >1$ we have:
$1 < a \le 2$.
=== badly worded earlier answer below ===
If $a > 1$ then $\log_a(x^2 + 4) \ge 2\iff a^{\log_a (x^2 + 4)}=x^2 +4 \ge a^2\iff \sqrt{x^2 + 4} \ge a$.  Now $\sqrt{x^2 + 4}$ can be as small, but no smaller than $2$ and can take on all values higher than $2$ so this can only be true for all $x$ (assuming $a > 1$) and will be true for all $x$ if $a \le 2$.
Or to put it another way.  If $a > 2$ then if $0 \le x < \sqrt{a^2 -4}$ we have $x^2 + 4 < a^2$ and $\log_a (x^2 + 4) < 2$.  But if $1 < a \le 2$ then $a^2 \le 4 \le x^2 + 4$ for all $x$ and $\log_2 (x^2 + 4) \ge 2$. 
And if $0< a < 1$ then $x^2 + 4 \ge 4 > 1$ so $\log_a x^2 + 4 < 0$ for all $x$.
And $\log_a$ is undefined for $a\le 0$ or $a = 0$.
so $\log_a (x^2 + 4) \ge 2 \iff 1 < a^2 < x^2 + 4$ and that is only true for all $x$ if $1< a \le 2$.
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