How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$? I guess it's easy, but I still need help. The inequality is
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
If you set $t=2^x$, then it becomes
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$

The set of solutions is
$$x \in (-\infty,-2) \cup \{1\}$$
which is not what I get.

My attempt
$$\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$$
This is undefined for $x=-2$, because the right hand side becomes: $\frac{1}{2^{-2+2}-1}=\frac{1}{2^0-1}=\frac{1}{1-1}=\frac{1}{0}$. Therefore,
$$x \neq -2$$
Now, let's set $t=2^x$.
$$\frac{1}{t+3} \geq \frac{1}{4t-1}$$
Multiply both sides with $(t+3)(4t-1)$.
We must split the inequality because we don't know is $(t+3)(4t-1)$ positive or negative. Question: what if it's zero? Should we consider that possibility too?
For $(t+3)(4t-1) > 0$:
$$4t-1 \geq t+3$$
$$3t \geq 4$$
$$t \geq \frac{4}{3}$$
$$2^x \geq \frac{4}{3}$$
$$x \geq \log_2 \left(\frac{4}{3}\right)$$
For $(t+3)(4t-1) < 0$:
$$4t-1 \leq t+3$$
$$3t \leq 4$$
$$t \leq \frac{4}{3}$$
$$2^x \leq \frac{4}{3}$$
$$x \leq \log_2 \left(\frac{4}{3}\right)$$
I think it's already obvious where and how I'm wrong, so I think I don't need to continue with this attempt. If I do, then please ask in the comment.

The given solution is clearly wrong on the positive side. For instance, take $x=2$, and 
$$
\frac17=\frac{1}{2^x+3}\geq\frac{1}{2^{x+2}-1}=\frac1{15}.
$$
Since $t=2^x$ you know that $t>0$. So $t+3>0$. Now if $4t-1>0$, everything is positive and you can multiply to get 
$$
4t-1\geq t+3,
$$
which simplifies to $3t\geq4$, or $t\geq4/3$. 
If, on the other hand, $4t-1<0$, the inequality will be reversed if we multiply by it. So we get 
$$
4t-1\leq t+3,
$$
which is $t\leq 4/3$. This was under the hypothesis $t<1/4$, so we get $t<1/4$. 
In summary, the inequality holds when $t<1/4$ and when $t\geq 4/3$. 
Now we translate to $x$. The equality $2^x=1/4$ gives $x=-2$. The equality $2^x=4/3$, gives $x=\log_24-\log_23=2-\log_23$. So the inequality holds for 
$$
x\in(-\infty,-2)\cup[2-\log_23,\infty)
$$
