How to solve $\frac{1}{2^x+3} \geq \frac{1}{2^{x+2}-1}$?

Question

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 $$\frac{1}{t+3} \geq \frac{1}{4t-1}$$

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

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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.

Answer 1

Continue from your efforts, $2^x = t$ already implies $t \gt 0$. $$\frac{1}{t+3}- \frac{1}{4t-1} \ge 0\\ \frac{3t-4}{(t+3)(4t-1)}\ge 0$$

Solving this inequality with the help of zero points of the three factors, gives $t\in (-3,1/4) \cup [4/3, \infty) $ but we also had $t\gt 0$. So final answer is $t \in (0,1/4) \cup [4/3, \infty)$.

So $x \in (-\infty, -2) \cup [2-\log_2(3), \infty)$

Answer 2

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 $$ x\in(-\infty,-2)\cup[2-\log_23,\infty) $$

Answer 3

Considering that $t>0$, the LHS is non-negative and the inequation certainly holds for $$t\lt\frac14.$$ Then for $t>\dfrac14$, the RHS is also non-negative and we need

$$t+3\le4t-1$$ or

$$t\ge\frac43.$$

Hence

$$x<\log_2\frac14\lor x\ge \log_2\frac43.$$